Mixing
Right identity and Right inverse implies a group
$begingroup$
Let $(G, *)$ be a semi-group. Suppose
- $ exists e in G$ such that $forall a in G, ae = a$;
- $forall a in G, exists a^{-1} in G$ such that $aa^{-1} = e$.
How can we prove that $(G,*)$ is a group?
group-theory
edited Sep 17 '11 at 14:46
Srivatsan
21k371126
asked Sep 17 '11 at 7:51
MohanMohan
6,0021353101
$endgroup$
|
show 2 more comments
$begingroup$
Let $(G, *)$ be a semi-group. Suppose
- $ exists e in G$ such that $forall a in G, ae = a$;
- $forall a in G, exists a^{-1} in G$ such that $aa^{-1} = e$.
How can we prove that $(G,*)$ is a group?
group-theory
edited Sep 17 '11 at 14:46
Srivatsan
21k371126
asked Sep 17 '11 at 7:51
MohanMohan
6,0021353101
$endgroup$
8
$begingroup$
In case you don't know: Right identity and Left inverse does not imply group.
$endgroup$
– j.p.
Sep 17 '11 at 9:45
13
$begingroup$
This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"?
$endgroup$
– Derek Holt
Sep 17 '11 at 15:31
4
$begingroup$
@Derek, I think the formulation is intended to be read as "Suppose there is $ein G$ such that 1 and 2.".
$endgroup$
– lhf
Sep 17 '11 at 15:52
2
$begingroup$
@lhf: Yes, that's the formally correct way to do it, and it also removes the ambiguity.
$endgroup$
– Derek Holt
Sep 17 '11 at 21:29
1
$begingroup$
@ThreeFx: Take a set with two elements and define $xcdot y=x$ (possibly $x=y$), which is associative and makes both elements to right identities. If you choose a right identity, then it's also the left inverse for both elements.
$endgroup$
– j.p.
Jan 9 '17 at 8:02
|
show 2 more comments
$begingroup$
Let $(G, *)$ be a semi-group. Suppose
- $ exists e in G$ such that $forall a in G, ae = a$;
- $forall a in G, exists a^{-1} in G$ such that $aa^{-1} = e$.
How can we prove that $(G,*)$ is a group?
group-theory
edited Sep 17 '11 at 14:46
Srivatsan
21k371126
asked Sep 17 '11 at 7:51
MohanMohan
6,0021353101
$endgroup$
Let $(G, *)$ be a semi-group. Suppose
- $ exists e in G$ such that $forall a in G, ae = a$;
- $forall a in G, exists a^{-1} in G$ such that $aa^{-1} = e$.
How can we prove that $(G,*)$ is a group?
group-theory
group-theory
edited Sep 17 '11 at 14:46
Srivatsan
21k371126
asked Sep 17 '11 at 7:51
MohanMohan
6,0021353101
edited Sep 17 '11 at 14:46
Srivatsan
21k371126
asked Sep 17 '11 at 7:51
MohanMohan
6,0021353101
edited Sep 17 '11 at 14:46
Srivatsan
21k371126
edited Sep 17 '11 at 14:46
Srivatsan
21k371126
edited Sep 17 '11 at 14:46
Srivatsan
21k371126
21k371126
asked Sep 17 '11 at 7:51
MohanMohan
6,0021353101
asked Sep 17 '11 at 7:51
MohanMohan
6,0021353101
asked Sep 17 '11 at 7:51
MohanMohan
6,0021353101
6,0021353101
8
$begingroup$
In case you don't know: Right identity and Left inverse does not imply group.
$endgroup$
– j.p.
Sep 17 '11 at 9:45
13
$begingroup$
This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"?
$endgroup$
– Derek Holt
Sep 17 '11 at 15:31
4
$begingroup$
@Derek, I think the formulation is intended to be read as "Suppose there is $ein G$ such that 1 and 2.".
$endgroup$
– lhf
Sep 17 '11 at 15:52
2
$begingroup$
@lhf: Yes, that's the formally correct way to do it, and it also removes the ambiguity.
$endgroup$
– Derek Holt
Sep 17 '11 at 21:29
1
$begingroup$
@ThreeFx: Take a set with two elements and define $xcdot y=x$ (possibly $x=y$), which is associative and makes both elements to right identities. If you choose a right identity, then it's also the left inverse for both elements.
$endgroup$
– j.p.
Jan 9 '17 at 8:02
|
show 2 more comments
8
$begingroup$
In case you don't know: Right identity and Left inverse does not imply group.
$endgroup$
– j.p.
Sep 17 '11 at 9:45
13
$begingroup$
This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"?
$endgroup$
– Derek Holt
Sep 17 '11 at 15:31
4
$begingroup$
@Derek, I think the formulation is intended to be read as "Suppose there is $ein G$ such that 1 and 2.".
$endgroup$
– lhf
Sep 17 '11 at 15:52
2
$begingroup$
@lhf: Yes, that's the formally correct way to do it, and it also removes the ambiguity.
$endgroup$
– Derek Holt
Sep 17 '11 at 21:29
1
$begingroup$
@ThreeFx: Take a set with two elements and define $xcdot y=x$ (possibly $x=y$), which is associative and makes both elements to right identities. If you choose a right identity, then it's also the left inverse for both elements.
$endgroup$
– j.p.
Jan 9 '17 at 8:02
8
8
$begingroup$
In case you don't know: Right identity and Left inverse does not imply group.
$endgroup$
– j.p.
Sep 17 '11 at 9:45
$begingroup$
In case you don't know: Right identity and Left inverse does not imply group.
$endgroup$
– j.p.
Sep 17 '11 at 9:45
13
13
$begingroup$
This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"?
$endgroup$
– Derek Holt
Sep 17 '11 at 15:31
$begingroup$
This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"?
$endgroup$
– Derek Holt
Sep 17 '11 at 15:31
4
4
$begingroup$
@Derek, I think the formulation is intended to be read as "Suppose there is $ein G$ such that 1 and 2.".
$endgroup$
– lhf
Sep 17 '11 at 15:52
$begingroup$
@Derek, I think the formulation is intended to be read as "Suppose there is $ein G$ such that 1 and 2.".
$endgroup$
– lhf
Sep 17 '11 at 15:52
2
2
$begingroup$
@lhf: Yes, that's the formally correct way to do it, and it also removes the ambiguity.
$endgroup$
– Derek Holt
Sep 17 '11 at 21:29
$begingroup$
@lhf: Yes, that's the formally correct way to do it, and it also removes the ambiguity.
$endgroup$
– Derek Holt
Sep 17 '11 at 21:29
1
1
$begingroup$
@ThreeFx: Take a set with two elements and define $xcdot y=x$ (possibly $x=y$), which is associative and makes both elements to right identities. If you choose a right identity, then it's also the left inverse for both elements.
$endgroup$
– j.p.
Jan 9 '17 at 8:02
$begingroup$
@ThreeFx: Take a set with two elements and define $xcdot y=x$ (possibly $x=y$), which is associative and makes both elements to right identities. If you choose a right identity, then it's also the left inverse for both elements.
$endgroup$
– j.p.
Jan 9 '17 at 8:02
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
I assume that (a) should read $exists ein G$ such that $ae=a$, $forall ain G$. For each $a in G$ we have
$$begin{align*}
(a^{-1})^{-1}a^{-1} &= e[(a^{-1})^{-1}a^{-1}]\
&= (aa^{-1})[(a^{-1})^{-1}a^{-1}]\
&= [(aa^{-1})(a^{-1})^{-1}]a^{-1}\
&= (a[a^{-1}(a^{-1})^{-1}])a^{-1}\
&= (ae)a^{-1}\
&= aa^{-1}.
end{align*}$$
Multiplying $(a^{-1})^{-1}a^{-1} = aa^{-1}$ on the right by $(a^{-1})^{-1}$ yields $$begin{align*}
(a^{-1})^{-1} &= (a^{-1})^{-1}e\
&= (a^{-1})^{-1}[a^{-1}(a^{-1})^{-1}]\
&= [(a^{-1})^{-1}a^{-1}](a^{-1})^{-1}\
&= (aa^{-1})(a^{-1})^{-1}\
&= a[a^{-1}(a^{-1})^{-1}]\
&= ae\
&= a,
end{align*}$$
so $a^{-1}a=e$ for all $a in G$.
Added: The foregoing obviously assumes that $e$ is a left identity, which was not given, and somehow none of us caught it at the time. Here is a corrected argument. For each $ain G$ we have $$a^{-1}=a^{-1}e=a^{-1}(aa^{-1})=(a^{-1}a)a^{-1};,$$ so $$e=a^{-1}(a^{-1})^{-1}=left((a^{-1}a)a^{-1}right)(a^{-1})^{-1}=(a^{-1}a)left(a^{-1}(a^{-1})^{-1}right)=(a^{-1}a)e=a^{-1}a;.$$
In other words, $a^{-1}$ is both a left as well as a right inverse for $a$. It follows that
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a;,$$
so $e$ is a left as well as a right identity for $G$. Now you can use the usual arguments to show that the identity and inverses are unique. (For example, if $e'$ were another identity, we’d have $e = ee' = e'$, because $e$ is a left identity and $e'$ is a right identity.)
answered Sep 17 '11 at 8:43
Brian M. ScottBrian M. Scott
459k38513916
$endgroup$
$begingroup$
It is not necessary to prove that $e$ and $cdot^{-1}$ are unique, that is not part of the (usual) axioms of a group.
$endgroup$
– Plop
Sep 17 '11 at 14:41
$begingroup$
@Plop Yes, I agree $cdot^{-1}$ need not be unique. But if $e$ is not unique, then how do we say that $a a^{-1} = a^{-1} a = e$? (This is a doubt, not a rhetorical question :-))
$endgroup$
– Srivatsan
Sep 17 '11 at 14:49
$begingroup$
@Sriv: Assume e is NOT unique. Then since by definition, G is closed under multiplication, the INVERSE of each element cannot be unique either.Otherwise,we would have "isolated" identities that do not result from the product of an element and it's inverse and G would not be closed under the product!Of course,that doesn't mean we can't have an algebraic structure like this-it just means the result is not a group. There is a new concept in algebra called a Beta group,in which there are infinitely many identities and inverses,but these are not groups per se.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:21
$begingroup$
@Math Thanks for the clarification and the term :).
$endgroup$
– Srivatsan
Sep 17 '11 at 16:23
$begingroup$
@Srivatsan What I meant was that from the usual axioms of a group, you can prove the uniqueness of $e$ and $cdot^{-1}$.
$endgroup$
– Plop
Sep 18 '11 at 10:51
|
show 3 more comments
$begingroup$
It is conceptually very simple that a right inverse is also a left inverse (when there is also a right identity). It follows from the axioms above in two steps:
1) Any element $a$ with the property $aa = a$ [i.e. idempotent] must be equal to the identity $e$ in the axioms, since in that case:
$$a = ae = a(aa^{-1}) = (aa)a^{-1} = aa^{-1} = e$$
This already proves the uniqueness of the [right] identity, since any identity by definition has the property of being idempotent.
2) By the axioms, for every element $a$ there is at least one right inverse element $a^{-1}$ such that $aa^{-1}=e$. Now we form the product of the same two elements in reverse order, namely $a^{-1}a$, to see if that product also equals the identity. If so, this right inverse is also a left inverse. We only need to show that $a^{-1}a$ is idempotent, and then its equality to $e$ follows from step 1:
$$[a^{-1}a][ a^{-1}a] = a^{-1}(a a^{-1})a = a^{-1}ea = a^{-1}a $$
3) It is now clear that the right identity is also a left identity. For any $a$:
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a$$
4) To show the uniqueness of the inverse:
Given any elements $a$ and $b$ such that $ab=e$, then
$$b = eb = a^{-1}ab = a^{-1}e = a^{-1}$$
Here, as above, the symbol $a^{-1}$ was first used to denote a representative right inverse of the element $a$. This inverse is now seen to be unique. Therefore, the symbol now signifies an operation of "inversion" which constitutes a single-valued function on the elements of the set.
See Richard A. Dean, “Elements of Abstract Algebra” (Wiley, 1967), pp 30-31.
answered Jul 23 '12 at 1:24
Douglas BilodeauDouglas Bilodeau
19113
$endgroup$
$begingroup$
and where on earth did we assume idempotency in the definition of a semigroup?
$endgroup$
– moldovean
Nov 6 '14 at 10:27
6
$begingroup$
@moldovean: We did not assume idempotency for any elements. It was shown that if an element is idempotent, then that element must be equal to any identity element that satisfies both 1. and 2. In a separate step, we then argued, that since any identity element (whether or not inverses exist for that element) is by definition idempotent all identity elements that exist are in fact equal to the identity element used in the proof. Ergo the identity element is unique.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:07
2
$begingroup$
This answer gets my vote for containing the most elegant proof for the problem, that I have come across.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:09
add a comment |
$begingroup$
This is stated with left identity and left inverse as Proposition 20.4 in the book Spindler: Abstract Algebra with Applications. Let me copy here the proof from this book (it should be easy for you to change it for the right instead of left):
Let $xin G$ be arbitrary. We want to show that the left inverse $x^{-1}$ is in fact also a right inverse. Let $y:=xx^{-1}$. Then
$$yy=(xx^{-1})(xx^{-1})=x(x^{-1}x)x^{-1}=x(ex^{-1})=xx^{-1}=y.$$
Hence
$$e=y^{-1}y=y^{-1}(yy)=(y^{-1}y)y=ey=y=xx^{-1},$$
i.e. $xx^{-1}=e$ which was what we wanted to show.
Now we prove that the left-neutral element $e$ is also a right-neutral element. Let $xin G$ be arbitrary; we want to establish that $xe=x$. Now
$$xe=x(x^{-1}x)=(xx^{-1})x=ex=x. $$
I googled a little and found out that several authors take this in fact as a definition of group, here are some of the first hits from google books when searching for "left inverse" "left identity" group:
- Robinson: A course in the theory of groups, p.2
- Gelbaum, Olmsted: Theorems and counterexamples in mathematics, p.1
- Sharma: Group Theory, p.14
edited Aug 16 '15 at 22:17
Ramashalanka
1309
answered Sep 17 '11 at 12:33
Martin SleziakMartin Sleziak
44.9k10120273
$endgroup$
2
$begingroup$
And of course,this is how both van der Waerden and Emil Artin define a group in their classic presentations of algebra. Later authors of textbooks generally found this "minimalist" method of defining a group far too tedious-so they assumed the stronger axioms and called it a day. A lot of classic algebra books-Herstien's TOPICS IN ALGEBRA,famously-use this as an exercise.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:24
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I assume that (a) should read $exists ein G$ such that $ae=a$, $forall ain G$. For each $a in G$ we have
$$begin{align*}
(a^{-1})^{-1}a^{-1} &= e[(a^{-1})^{-1}a^{-1}]\
&= (aa^{-1})[(a^{-1})^{-1}a^{-1}]\
&= [(aa^{-1})(a^{-1})^{-1}]a^{-1}\
&= (a[a^{-1}(a^{-1})^{-1}])a^{-1}\
&= (ae)a^{-1}\
&= aa^{-1}.
end{align*}$$
Multiplying $(a^{-1})^{-1}a^{-1} = aa^{-1}$ on the right by $(a^{-1})^{-1}$ yields $$begin{align*}
(a^{-1})^{-1} &= (a^{-1})^{-1}e\
&= (a^{-1})^{-1}[a^{-1}(a^{-1})^{-1}]\
&= [(a^{-1})^{-1}a^{-1}](a^{-1})^{-1}\
&= (aa^{-1})(a^{-1})^{-1}\
&= a[a^{-1}(a^{-1})^{-1}]\
&= ae\
&= a,
end{align*}$$
so $a^{-1}a=e$ for all $a in G$.
Added: The foregoing obviously assumes that $e$ is a left identity, which was not given, and somehow none of us caught it at the time. Here is a corrected argument. For each $ain G$ we have $$a^{-1}=a^{-1}e=a^{-1}(aa^{-1})=(a^{-1}a)a^{-1};,$$ so $$e=a^{-1}(a^{-1})^{-1}=left((a^{-1}a)a^{-1}right)(a^{-1})^{-1}=(a^{-1}a)left(a^{-1}(a^{-1})^{-1}right)=(a^{-1}a)e=a^{-1}a;.$$
In other words, $a^{-1}$ is both a left as well as a right inverse for $a$. It follows that
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a;,$$
so $e$ is a left as well as a right identity for $G$. Now you can use the usual arguments to show that the identity and inverses are unique. (For example, if $e'$ were another identity, we’d have $e = ee' = e'$, because $e$ is a left identity and $e'$ is a right identity.)
answered Sep 17 '11 at 8:43
Brian M. ScottBrian M. Scott
459k38513916
$endgroup$
$begingroup$
It is not necessary to prove that $e$ and $cdot^{-1}$ are unique, that is not part of the (usual) axioms of a group.
$endgroup$
– Plop
Sep 17 '11 at 14:41
$begingroup$
@Plop Yes, I agree $cdot^{-1}$ need not be unique. But if $e$ is not unique, then how do we say that $a a^{-1} = a^{-1} a = e$? (This is a doubt, not a rhetorical question :-))
$endgroup$
– Srivatsan
Sep 17 '11 at 14:49
$begingroup$
@Sriv: Assume e is NOT unique. Then since by definition, G is closed under multiplication, the INVERSE of each element cannot be unique either.Otherwise,we would have "isolated" identities that do not result from the product of an element and it's inverse and G would not be closed under the product!Of course,that doesn't mean we can't have an algebraic structure like this-it just means the result is not a group. There is a new concept in algebra called a Beta group,in which there are infinitely many identities and inverses,but these are not groups per se.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:21
$begingroup$
@Math Thanks for the clarification and the term :).
$endgroup$
– Srivatsan
Sep 17 '11 at 16:23
$begingroup$
@Srivatsan What I meant was that from the usual axioms of a group, you can prove the uniqueness of $e$ and $cdot^{-1}$.
$endgroup$
– Plop
Sep 18 '11 at 10:51
|
show 3 more comments
$begingroup$
I assume that (a) should read $exists ein G$ such that $ae=a$, $forall ain G$. For each $a in G$ we have
$$begin{align*}
(a^{-1})^{-1}a^{-1} &= e[(a^{-1})^{-1}a^{-1}]\
&= (aa^{-1})[(a^{-1})^{-1}a^{-1}]\
&= [(aa^{-1})(a^{-1})^{-1}]a^{-1}\
&= (a[a^{-1}(a^{-1})^{-1}])a^{-1}\
&= (ae)a^{-1}\
&= aa^{-1}.
end{align*}$$
Multiplying $(a^{-1})^{-1}a^{-1} = aa^{-1}$ on the right by $(a^{-1})^{-1}$ yields $$begin{align*}
(a^{-1})^{-1} &= (a^{-1})^{-1}e\
&= (a^{-1})^{-1}[a^{-1}(a^{-1})^{-1}]\
&= [(a^{-1})^{-1}a^{-1}](a^{-1})^{-1}\
&= (aa^{-1})(a^{-1})^{-1}\
&= a[a^{-1}(a^{-1})^{-1}]\
&= ae\
&= a,
end{align*}$$
so $a^{-1}a=e$ for all $a in G$.
Added: The foregoing obviously assumes that $e$ is a left identity, which was not given, and somehow none of us caught it at the time. Here is a corrected argument. For each $ain G$ we have $$a^{-1}=a^{-1}e=a^{-1}(aa^{-1})=(a^{-1}a)a^{-1};,$$ so $$e=a^{-1}(a^{-1})^{-1}=left((a^{-1}a)a^{-1}right)(a^{-1})^{-1}=(a^{-1}a)left(a^{-1}(a^{-1})^{-1}right)=(a^{-1}a)e=a^{-1}a;.$$
In other words, $a^{-1}$ is both a left as well as a right inverse for $a$. It follows that
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a;,$$
so $e$ is a left as well as a right identity for $G$. Now you can use the usual arguments to show that the identity and inverses are unique. (For example, if $e'$ were another identity, we’d have $e = ee' = e'$, because $e$ is a left identity and $e'$ is a right identity.)
answered Sep 17 '11 at 8:43
Brian M. ScottBrian M. Scott
459k38513916
$endgroup$
$begingroup$
It is not necessary to prove that $e$ and $cdot^{-1}$ are unique, that is not part of the (usual) axioms of a group.
$endgroup$
– Plop
Sep 17 '11 at 14:41
$begingroup$
@Plop Yes, I agree $cdot^{-1}$ need not be unique. But if $e$ is not unique, then how do we say that $a a^{-1} = a^{-1} a = e$? (This is a doubt, not a rhetorical question :-))
$endgroup$
– Srivatsan
Sep 17 '11 at 14:49
$begingroup$
@Sriv: Assume e is NOT unique. Then since by definition, G is closed under multiplication, the INVERSE of each element cannot be unique either.Otherwise,we would have "isolated" identities that do not result from the product of an element and it's inverse and G would not be closed under the product!Of course,that doesn't mean we can't have an algebraic structure like this-it just means the result is not a group. There is a new concept in algebra called a Beta group,in which there are infinitely many identities and inverses,but these are not groups per se.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:21
$begingroup$
@Math Thanks for the clarification and the term :).
$endgroup$
– Srivatsan
Sep 17 '11 at 16:23
$begingroup$
@Srivatsan What I meant was that from the usual axioms of a group, you can prove the uniqueness of $e$ and $cdot^{-1}$.
$endgroup$
– Plop
Sep 18 '11 at 10:51
|
show 3 more comments
$begingroup$
I assume that (a) should read $exists ein G$ such that $ae=a$, $forall ain G$. For each $a in G$ we have
$$begin{align*}
(a^{-1})^{-1}a^{-1} &= e[(a^{-1})^{-1}a^{-1}]\
&= (aa^{-1})[(a^{-1})^{-1}a^{-1}]\
&= [(aa^{-1})(a^{-1})^{-1}]a^{-1}\
&= (a[a^{-1}(a^{-1})^{-1}])a^{-1}\
&= (ae)a^{-1}\
&= aa^{-1}.
end{align*}$$
Multiplying $(a^{-1})^{-1}a^{-1} = aa^{-1}$ on the right by $(a^{-1})^{-1}$ yields $$begin{align*}
(a^{-1})^{-1} &= (a^{-1})^{-1}e\
&= (a^{-1})^{-1}[a^{-1}(a^{-1})^{-1}]\
&= [(a^{-1})^{-1}a^{-1}](a^{-1})^{-1}\
&= (aa^{-1})(a^{-1})^{-1}\
&= a[a^{-1}(a^{-1})^{-1}]\
&= ae\
&= a,
end{align*}$$
so $a^{-1}a=e$ for all $a in G$.
Added: The foregoing obviously assumes that $e$ is a left identity, which was not given, and somehow none of us caught it at the time. Here is a corrected argument. For each $ain G$ we have $$a^{-1}=a^{-1}e=a^{-1}(aa^{-1})=(a^{-1}a)a^{-1};,$$ so $$e=a^{-1}(a^{-1})^{-1}=left((a^{-1}a)a^{-1}right)(a^{-1})^{-1}=(a^{-1}a)left(a^{-1}(a^{-1})^{-1}right)=(a^{-1}a)e=a^{-1}a;.$$
In other words, $a^{-1}$ is both a left as well as a right inverse for $a$. It follows that
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a;,$$
so $e$ is a left as well as a right identity for $G$. Now you can use the usual arguments to show that the identity and inverses are unique. (For example, if $e'$ were another identity, we’d have $e = ee' = e'$, because $e$ is a left identity and $e'$ is a right identity.)
answered Sep 17 '11 at 8:43
Brian M. ScottBrian M. Scott
459k38513916
$endgroup$
I assume that (a) should read $exists ein G$ such that $ae=a$, $forall ain G$. For each $a in G$ we have
$$begin{align*}
(a^{-1})^{-1}a^{-1} &= e[(a^{-1})^{-1}a^{-1}]\
&= (aa^{-1})[(a^{-1})^{-1}a^{-1}]\
&= [(aa^{-1})(a^{-1})^{-1}]a^{-1}\
&= (a[a^{-1}(a^{-1})^{-1}])a^{-1}\
&= (ae)a^{-1}\
&= aa^{-1}.
end{align*}$$
Multiplying $(a^{-1})^{-1}a^{-1} = aa^{-1}$ on the right by $(a^{-1})^{-1}$ yields $$begin{align*}
(a^{-1})^{-1} &= (a^{-1})^{-1}e\
&= (a^{-1})^{-1}[a^{-1}(a^{-1})^{-1}]\
&= [(a^{-1})^{-1}a^{-1}](a^{-1})^{-1}\
&= (aa^{-1})(a^{-1})^{-1}\
&= a[a^{-1}(a^{-1})^{-1}]\
&= ae\
&= a,
end{align*}$$
so $a^{-1}a=e$ for all $a in G$.
Added: The foregoing obviously assumes that $e$ is a left identity, which was not given, and somehow none of us caught it at the time. Here is a corrected argument. For each $ain G$ we have $$a^{-1}=a^{-1}e=a^{-1}(aa^{-1})=(a^{-1}a)a^{-1};,$$ so $$e=a^{-1}(a^{-1})^{-1}=left((a^{-1}a)a^{-1}right)(a^{-1})^{-1}=(a^{-1}a)left(a^{-1}(a^{-1})^{-1}right)=(a^{-1}a)e=a^{-1}a;.$$
In other words, $a^{-1}$ is both a left as well as a right inverse for $a$. It follows that
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a;,$$
so $e$ is a left as well as a right identity for $G$. Now you can use the usual arguments to show that the identity and inverses are unique. (For example, if $e'$ were another identity, we’d have $e = ee' = e'$, because $e$ is a left identity and $e'$ is a right identity.)
answered Sep 17 '11 at 8:43
Brian M. ScottBrian M. Scott
459k38513916
edited Dec 3 '13 at 5:02
answered Sep 17 '11 at 8:43
Brian M. ScottBrian M. Scott
459k38513916
answered Sep 17 '11 at 8:43
Brian M. ScottBrian M. Scott
459k38513916
answered Sep 17 '11 at 8:43
Brian M. ScottBrian M. Scott
459k38513916
459k38513916
$begingroup$
It is not necessary to prove that $e$ and $cdot^{-1}$ are unique, that is not part of the (usual) axioms of a group.
$endgroup$
– Plop
Sep 17 '11 at 14:41
$begingroup$
@Plop Yes, I agree $cdot^{-1}$ need not be unique. But if $e$ is not unique, then how do we say that $a a^{-1} = a^{-1} a = e$? (This is a doubt, not a rhetorical question :-))
$endgroup$
– Srivatsan
Sep 17 '11 at 14:49
$begingroup$
@Sriv: Assume e is NOT unique. Then since by definition, G is closed under multiplication, the INVERSE of each element cannot be unique either.Otherwise,we would have "isolated" identities that do not result from the product of an element and it's inverse and G would not be closed under the product!Of course,that doesn't mean we can't have an algebraic structure like this-it just means the result is not a group. There is a new concept in algebra called a Beta group,in which there are infinitely many identities and inverses,but these are not groups per se.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:21
$begingroup$
@Math Thanks for the clarification and the term :).
$endgroup$
– Srivatsan
Sep 17 '11 at 16:23
$begingroup$
@Srivatsan What I meant was that from the usual axioms of a group, you can prove the uniqueness of $e$ and $cdot^{-1}$.
$endgroup$
– Plop
Sep 18 '11 at 10:51
|
show 3 more comments
$begingroup$
It is not necessary to prove that $e$ and $cdot^{-1}$ are unique, that is not part of the (usual) axioms of a group.
$endgroup$
– Plop
Sep 17 '11 at 14:41
$begingroup$
@Plop Yes, I agree $cdot^{-1}$ need not be unique. But if $e$ is not unique, then how do we say that $a a^{-1} = a^{-1} a = e$? (This is a doubt, not a rhetorical question :-))
$endgroup$
– Srivatsan
Sep 17 '11 at 14:49
$begingroup$
@Sriv: Assume e is NOT unique. Then since by definition, G is closed under multiplication, the INVERSE of each element cannot be unique either.Otherwise,we would have "isolated" identities that do not result from the product of an element and it's inverse and G would not be closed under the product!Of course,that doesn't mean we can't have an algebraic structure like this-it just means the result is not a group. There is a new concept in algebra called a Beta group,in which there are infinitely many identities and inverses,but these are not groups per se.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:21
$begingroup$
@Math Thanks for the clarification and the term :).
$endgroup$
– Srivatsan
Sep 17 '11 at 16:23
$begingroup$
@Srivatsan What I meant was that from the usual axioms of a group, you can prove the uniqueness of $e$ and $cdot^{-1}$.
$endgroup$
– Plop
Sep 18 '11 at 10:51
$begingroup$
It is not necessary to prove that $e$ and $cdot^{-1}$ are unique, that is not part of the (usual) axioms of a group.
$endgroup$
– Plop
Sep 17 '11 at 14:41
$begingroup$
It is not necessary to prove that $e$ and $cdot^{-1}$ are unique, that is not part of the (usual) axioms of a group.
$endgroup$
– Plop
Sep 17 '11 at 14:41
$begingroup$
@Plop Yes, I agree $cdot^{-1}$ need not be unique. But if $e$ is not unique, then how do we say that $a a^{-1} = a^{-1} a = e$? (This is a doubt, not a rhetorical question :-))
$endgroup$
– Srivatsan
Sep 17 '11 at 14:49
$begingroup$
@Plop Yes, I agree $cdot^{-1}$ need not be unique. But if $e$ is not unique, then how do we say that $a a^{-1} = a^{-1} a = e$? (This is a doubt, not a rhetorical question :-))
$endgroup$
– Srivatsan
Sep 17 '11 at 14:49
$begingroup$
@Sriv: Assume e is NOT unique. Then since by definition, G is closed under multiplication, the INVERSE of each element cannot be unique either.Otherwise,we would have "isolated" identities that do not result from the product of an element and it's inverse and G would not be closed under the product!Of course,that doesn't mean we can't have an algebraic structure like this-it just means the result is not a group. There is a new concept in algebra called a Beta group,in which there are infinitely many identities and inverses,but these are not groups per se.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:21
$begingroup$
@Sriv: Assume e is NOT unique. Then since by definition, G is closed under multiplication, the INVERSE of each element cannot be unique either.Otherwise,we would have "isolated" identities that do not result from the product of an element and it's inverse and G would not be closed under the product!Of course,that doesn't mean we can't have an algebraic structure like this-it just means the result is not a group. There is a new concept in algebra called a Beta group,in which there are infinitely many identities and inverses,but these are not groups per se.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:21
$begingroup$
@Math Thanks for the clarification and the term :).
$endgroup$
– Srivatsan
Sep 17 '11 at 16:23
$begingroup$
@Math Thanks for the clarification and the term :).
$endgroup$
– Srivatsan
Sep 17 '11 at 16:23
$begingroup$
@Srivatsan What I meant was that from the usual axioms of a group, you can prove the uniqueness of $e$ and $cdot^{-1}$.
$endgroup$
– Plop
Sep 18 '11 at 10:51
$begingroup$
@Srivatsan What I meant was that from the usual axioms of a group, you can prove the uniqueness of $e$ and $cdot^{-1}$.
$endgroup$
– Plop
Sep 18 '11 at 10:51
|
show 3 more comments
$begingroup$
It is conceptually very simple that a right inverse is also a left inverse (when there is also a right identity). It follows from the axioms above in two steps:
1) Any element $a$ with the property $aa = a$ [i.e. idempotent] must be equal to the identity $e$ in the axioms, since in that case:
$$a = ae = a(aa^{-1}) = (aa)a^{-1} = aa^{-1} = e$$
This already proves the uniqueness of the [right] identity, since any identity by definition has the property of being idempotent.
2) By the axioms, for every element $a$ there is at least one right inverse element $a^{-1}$ such that $aa^{-1}=e$. Now we form the product of the same two elements in reverse order, namely $a^{-1}a$, to see if that product also equals the identity. If so, this right inverse is also a left inverse. We only need to show that $a^{-1}a$ is idempotent, and then its equality to $e$ follows from step 1:
$$[a^{-1}a][ a^{-1}a] = a^{-1}(a a^{-1})a = a^{-1}ea = a^{-1}a $$
3) It is now clear that the right identity is also a left identity. For any $a$:
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a$$
4) To show the uniqueness of the inverse:
Given any elements $a$ and $b$ such that $ab=e$, then
$$b = eb = a^{-1}ab = a^{-1}e = a^{-1}$$
Here, as above, the symbol $a^{-1}$ was first used to denote a representative right inverse of the element $a$. This inverse is now seen to be unique. Therefore, the symbol now signifies an operation of "inversion" which constitutes a single-valued function on the elements of the set.
See Richard A. Dean, “Elements of Abstract Algebra” (Wiley, 1967), pp 30-31.
answered Jul 23 '12 at 1:24
Douglas BilodeauDouglas Bilodeau
19113
$endgroup$
$begingroup$
and where on earth did we assume idempotency in the definition of a semigroup?
$endgroup$
– moldovean
Nov 6 '14 at 10:27
6
$begingroup$
@moldovean: We did not assume idempotency for any elements. It was shown that if an element is idempotent, then that element must be equal to any identity element that satisfies both 1. and 2. In a separate step, we then argued, that since any identity element (whether or not inverses exist for that element) is by definition idempotent all identity elements that exist are in fact equal to the identity element used in the proof. Ergo the identity element is unique.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:07
2
$begingroup$
This answer gets my vote for containing the most elegant proof for the problem, that I have come across.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:09
add a comment |
$begingroup$
It is conceptually very simple that a right inverse is also a left inverse (when there is also a right identity). It follows from the axioms above in two steps:
1) Any element $a$ with the property $aa = a$ [i.e. idempotent] must be equal to the identity $e$ in the axioms, since in that case:
$$a = ae = a(aa^{-1}) = (aa)a^{-1} = aa^{-1} = e$$
This already proves the uniqueness of the [right] identity, since any identity by definition has the property of being idempotent.
2) By the axioms, for every element $a$ there is at least one right inverse element $a^{-1}$ such that $aa^{-1}=e$. Now we form the product of the same two elements in reverse order, namely $a^{-1}a$, to see if that product also equals the identity. If so, this right inverse is also a left inverse. We only need to show that $a^{-1}a$ is idempotent, and then its equality to $e$ follows from step 1:
$$[a^{-1}a][ a^{-1}a] = a^{-1}(a a^{-1})a = a^{-1}ea = a^{-1}a $$
3) It is now clear that the right identity is also a left identity. For any $a$:
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a$$
4) To show the uniqueness of the inverse:
Given any elements $a$ and $b$ such that $ab=e$, then
$$b = eb = a^{-1}ab = a^{-1}e = a^{-1}$$
Here, as above, the symbol $a^{-1}$ was first used to denote a representative right inverse of the element $a$. This inverse is now seen to be unique. Therefore, the symbol now signifies an operation of "inversion" which constitutes a single-valued function on the elements of the set.
See Richard A. Dean, “Elements of Abstract Algebra” (Wiley, 1967), pp 30-31.
answered Jul 23 '12 at 1:24
Douglas BilodeauDouglas Bilodeau
19113
$endgroup$
$begingroup$
and where on earth did we assume idempotency in the definition of a semigroup?
$endgroup$
– moldovean
Nov 6 '14 at 10:27
6
$begingroup$
@moldovean: We did not assume idempotency for any elements. It was shown that if an element is idempotent, then that element must be equal to any identity element that satisfies both 1. and 2. In a separate step, we then argued, that since any identity element (whether or not inverses exist for that element) is by definition idempotent all identity elements that exist are in fact equal to the identity element used in the proof. Ergo the identity element is unique.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:07
2
$begingroup$
This answer gets my vote for containing the most elegant proof for the problem, that I have come across.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:09
add a comment |
$begingroup$
It is conceptually very simple that a right inverse is also a left inverse (when there is also a right identity). It follows from the axioms above in two steps:
1) Any element $a$ with the property $aa = a$ [i.e. idempotent] must be equal to the identity $e$ in the axioms, since in that case:
$$a = ae = a(aa^{-1}) = (aa)a^{-1} = aa^{-1} = e$$
This already proves the uniqueness of the [right] identity, since any identity by definition has the property of being idempotent.
2) By the axioms, for every element $a$ there is at least one right inverse element $a^{-1}$ such that $aa^{-1}=e$. Now we form the product of the same two elements in reverse order, namely $a^{-1}a$, to see if that product also equals the identity. If so, this right inverse is also a left inverse. We only need to show that $a^{-1}a$ is idempotent, and then its equality to $e$ follows from step 1:
$$[a^{-1}a][ a^{-1}a] = a^{-1}(a a^{-1})a = a^{-1}ea = a^{-1}a $$
3) It is now clear that the right identity is also a left identity. For any $a$:
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a$$
4) To show the uniqueness of the inverse:
Given any elements $a$ and $b$ such that $ab=e$, then
$$b = eb = a^{-1}ab = a^{-1}e = a^{-1}$$
Here, as above, the symbol $a^{-1}$ was first used to denote a representative right inverse of the element $a$. This inverse is now seen to be unique. Therefore, the symbol now signifies an operation of "inversion" which constitutes a single-valued function on the elements of the set.
See Richard A. Dean, “Elements of Abstract Algebra” (Wiley, 1967), pp 30-31.
answered Jul 23 '12 at 1:24
Douglas BilodeauDouglas Bilodeau
19113
$endgroup$
It is conceptually very simple that a right inverse is also a left inverse (when there is also a right identity). It follows from the axioms above in two steps:
1) Any element $a$ with the property $aa = a$ [i.e. idempotent] must be equal to the identity $e$ in the axioms, since in that case:
$$a = ae = a(aa^{-1}) = (aa)a^{-1} = aa^{-1} = e$$
This already proves the uniqueness of the [right] identity, since any identity by definition has the property of being idempotent.
2) By the axioms, for every element $a$ there is at least one right inverse element $a^{-1}$ such that $aa^{-1}=e$. Now we form the product of the same two elements in reverse order, namely $a^{-1}a$, to see if that product also equals the identity. If so, this right inverse is also a left inverse. We only need to show that $a^{-1}a$ is idempotent, and then its equality to $e$ follows from step 1:
$$[a^{-1}a][ a^{-1}a] = a^{-1}(a a^{-1})a = a^{-1}ea = a^{-1}a $$
3) It is now clear that the right identity is also a left identity. For any $a$:
$$ea = (aa^{-1})a = a(a^{-1}a) = ae = a$$
4) To show the uniqueness of the inverse:
Given any elements $a$ and $b$ such that $ab=e$, then
$$b = eb = a^{-1}ab = a^{-1}e = a^{-1}$$
Here, as above, the symbol $a^{-1}$ was first used to denote a representative right inverse of the element $a$. This inverse is now seen to be unique. Therefore, the symbol now signifies an operation of "inversion" which constitutes a single-valued function on the elements of the set.
See Richard A. Dean, “Elements of Abstract Algebra” (Wiley, 1967), pp 30-31.
answered Jul 23 '12 at 1:24
Douglas BilodeauDouglas Bilodeau
19113
edited Jul 24 '12 at 16:36
answered Jul 23 '12 at 1:24
Douglas BilodeauDouglas Bilodeau
19113
answered Jul 23 '12 at 1:24
Douglas BilodeauDouglas Bilodeau
19113
answered Jul 23 '12 at 1:24
Douglas BilodeauDouglas Bilodeau
19113
19113
$begingroup$
and where on earth did we assume idempotency in the definition of a semigroup?
$endgroup$
– moldovean
Nov 6 '14 at 10:27
6
$begingroup$
@moldovean: We did not assume idempotency for any elements. It was shown that if an element is idempotent, then that element must be equal to any identity element that satisfies both 1. and 2. In a separate step, we then argued, that since any identity element (whether or not inverses exist for that element) is by definition idempotent all identity elements that exist are in fact equal to the identity element used in the proof. Ergo the identity element is unique.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:07
2
$begingroup$
This answer gets my vote for containing the most elegant proof for the problem, that I have come across.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:09
add a comment |
$begingroup$
and where on earth did we assume idempotency in the definition of a semigroup?
$endgroup$
– moldovean
Nov 6 '14 at 10:27
6
$begingroup$
@moldovean: We did not assume idempotency for any elements. It was shown that if an element is idempotent, then that element must be equal to any identity element that satisfies both 1. and 2. In a separate step, we then argued, that since any identity element (whether or not inverses exist for that element) is by definition idempotent all identity elements that exist are in fact equal to the identity element used in the proof. Ergo the identity element is unique.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:07
2
$begingroup$
This answer gets my vote for containing the most elegant proof for the problem, that I have come across.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:09
$begingroup$
and where on earth did we assume idempotency in the definition of a semigroup?
$endgroup$
– moldovean
Nov 6 '14 at 10:27
$begingroup$
and where on earth did we assume idempotency in the definition of a semigroup?
$endgroup$
– moldovean
Nov 6 '14 at 10:27
6
6
$begingroup$
@moldovean: We did not assume idempotency for any elements. It was shown that if an element is idempotent, then that element must be equal to any identity element that satisfies both 1. and 2. In a separate step, we then argued, that since any identity element (whether or not inverses exist for that element) is by definition idempotent all identity elements that exist are in fact equal to the identity element used in the proof. Ergo the identity element is unique.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:07
$begingroup$
@moldovean: We did not assume idempotency for any elements. It was shown that if an element is idempotent, then that element must be equal to any identity element that satisfies both 1. and 2. In a separate step, we then argued, that since any identity element (whether or not inverses exist for that element) is by definition idempotent all identity elements that exist are in fact equal to the identity element used in the proof. Ergo the identity element is unique.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:07
2
2
$begingroup$
This answer gets my vote for containing the most elegant proof for the problem, that I have come across.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:09
$begingroup$
This answer gets my vote for containing the most elegant proof for the problem, that I have come across.
$endgroup$
– Psachnodaimonia
Mar 21 '15 at 12:09
add a comment |
$begingroup$
This is stated with left identity and left inverse as Proposition 20.4 in the book Spindler: Abstract Algebra with Applications. Let me copy here the proof from this book (it should be easy for you to change it for the right instead of left):
Let $xin G$ be arbitrary. We want to show that the left inverse $x^{-1}$ is in fact also a right inverse. Let $y:=xx^{-1}$. Then
$$yy=(xx^{-1})(xx^{-1})=x(x^{-1}x)x^{-1}=x(ex^{-1})=xx^{-1}=y.$$
Hence
$$e=y^{-1}y=y^{-1}(yy)=(y^{-1}y)y=ey=y=xx^{-1},$$
i.e. $xx^{-1}=e$ which was what we wanted to show.
Now we prove that the left-neutral element $e$ is also a right-neutral element. Let $xin G$ be arbitrary; we want to establish that $xe=x$. Now
$$xe=x(x^{-1}x)=(xx^{-1})x=ex=x. $$
I googled a little and found out that several authors take this in fact as a definition of group, here are some of the first hits from google books when searching for "left inverse" "left identity" group:
- Robinson: A course in the theory of groups, p.2
- Gelbaum, Olmsted: Theorems and counterexamples in mathematics, p.1
- Sharma: Group Theory, p.14
edited Aug 16 '15 at 22:17
Ramashalanka
1309
answered Sep 17 '11 at 12:33
Martin SleziakMartin Sleziak
44.9k10120273
$endgroup$
2
$begingroup$
And of course,this is how both van der Waerden and Emil Artin define a group in their classic presentations of algebra. Later authors of textbooks generally found this "minimalist" method of defining a group far too tedious-so they assumed the stronger axioms and called it a day. A lot of classic algebra books-Herstien's TOPICS IN ALGEBRA,famously-use this as an exercise.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:24
add a comment |
$begingroup$
This is stated with left identity and left inverse as Proposition 20.4 in the book Spindler: Abstract Algebra with Applications. Let me copy here the proof from this book (it should be easy for you to change it for the right instead of left):
Let $xin G$ be arbitrary. We want to show that the left inverse $x^{-1}$ is in fact also a right inverse. Let $y:=xx^{-1}$. Then
$$yy=(xx^{-1})(xx^{-1})=x(x^{-1}x)x^{-1}=x(ex^{-1})=xx^{-1}=y.$$
Hence
$$e=y^{-1}y=y^{-1}(yy)=(y^{-1}y)y=ey=y=xx^{-1},$$
i.e. $xx^{-1}=e$ which was what we wanted to show.
Now we prove that the left-neutral element $e$ is also a right-neutral element. Let $xin G$ be arbitrary; we want to establish that $xe=x$. Now
$$xe=x(x^{-1}x)=(xx^{-1})x=ex=x. $$
I googled a little and found out that several authors take this in fact as a definition of group, here are some of the first hits from google books when searching for "left inverse" "left identity" group:
- Robinson: A course in the theory of groups, p.2
- Gelbaum, Olmsted: Theorems and counterexamples in mathematics, p.1
- Sharma: Group Theory, p.14
edited Aug 16 '15 at 22:17
Ramashalanka
1309
answered Sep 17 '11 at 12:33
Martin SleziakMartin Sleziak
44.9k10120273
$endgroup$
2
$begingroup$
And of course,this is how both van der Waerden and Emil Artin define a group in their classic presentations of algebra. Later authors of textbooks generally found this "minimalist" method of defining a group far too tedious-so they assumed the stronger axioms and called it a day. A lot of classic algebra books-Herstien's TOPICS IN ALGEBRA,famously-use this as an exercise.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:24
add a comment |
$begingroup$
This is stated with left identity and left inverse as Proposition 20.4 in the book Spindler: Abstract Algebra with Applications. Let me copy here the proof from this book (it should be easy for you to change it for the right instead of left):
Let $xin G$ be arbitrary. We want to show that the left inverse $x^{-1}$ is in fact also a right inverse. Let $y:=xx^{-1}$. Then
$$yy=(xx^{-1})(xx^{-1})=x(x^{-1}x)x^{-1}=x(ex^{-1})=xx^{-1}=y.$$
Hence
$$e=y^{-1}y=y^{-1}(yy)=(y^{-1}y)y=ey=y=xx^{-1},$$
i.e. $xx^{-1}=e$ which was what we wanted to show.
Now we prove that the left-neutral element $e$ is also a right-neutral element. Let $xin G$ be arbitrary; we want to establish that $xe=x$. Now
$$xe=x(x^{-1}x)=(xx^{-1})x=ex=x. $$
I googled a little and found out that several authors take this in fact as a definition of group, here are some of the first hits from google books when searching for "left inverse" "left identity" group:
- Robinson: A course in the theory of groups, p.2
- Gelbaum, Olmsted: Theorems and counterexamples in mathematics, p.1
- Sharma: Group Theory, p.14
edited Aug 16 '15 at 22:17
Ramashalanka
1309
answered Sep 17 '11 at 12:33
Martin SleziakMartin Sleziak
44.9k10120273
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This is stated with left identity and left inverse as Proposition 20.4 in the book Spindler: Abstract Algebra with Applications. Let me copy here the proof from this book (it should be easy for you to change it for the right instead of left):
Let $xin G$ be arbitrary. We want to show that the left inverse $x^{-1}$ is in fact also a right inverse. Let $y:=xx^{-1}$. Then
$$yy=(xx^{-1})(xx^{-1})=x(x^{-1}x)x^{-1}=x(ex^{-1})=xx^{-1}=y.$$
Hence
$$e=y^{-1}y=y^{-1}(yy)=(y^{-1}y)y=ey=y=xx^{-1},$$
i.e. $xx^{-1}=e$ which was what we wanted to show.
Now we prove that the left-neutral element $e$ is also a right-neutral element. Let $xin G$ be arbitrary; we want to establish that $xe=x$. Now
$$xe=x(x^{-1}x)=(xx^{-1})x=ex=x. $$
I googled a little and found out that several authors take this in fact as a definition of group, here are some of the first hits from google books when searching for "left inverse" "left identity" group:
- Robinson: A course in the theory of groups, p.2
- Gelbaum, Olmsted: Theorems and counterexamples in mathematics, p.1
- Sharma: Group Theory, p.14
edited Aug 16 '15 at 22:17
Ramashalanka
1309
answered Sep 17 '11 at 12:33
Martin SleziakMartin Sleziak
44.9k10120273
edited Aug 16 '15 at 22:17
Ramashalanka
1309
edited Aug 16 '15 at 22:17
Ramashalanka
1309
edited Aug 16 '15 at 22:17
Ramashalanka
1309
1309
answered Sep 17 '11 at 12:33
Martin SleziakMartin Sleziak
44.9k10120273
answered Sep 17 '11 at 12:33
Martin SleziakMartin Sleziak
44.9k10120273
answered Sep 17 '11 at 12:33
Martin SleziakMartin Sleziak
44.9k10120273
44.9k10120273
2
$begingroup$
And of course,this is how both van der Waerden and Emil Artin define a group in their classic presentations of algebra. Later authors of textbooks generally found this "minimalist" method of defining a group far too tedious-so they assumed the stronger axioms and called it a day. A lot of classic algebra books-Herstien's TOPICS IN ALGEBRA,famously-use this as an exercise.
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– Mathemagician1234
Sep 17 '11 at 16:24
add a comment |
2
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And of course,this is how both van der Waerden and Emil Artin define a group in their classic presentations of algebra. Later authors of textbooks generally found this "minimalist" method of defining a group far too tedious-so they assumed the stronger axioms and called it a day. A lot of classic algebra books-Herstien's TOPICS IN ALGEBRA,famously-use this as an exercise.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:24
2
2
$begingroup$
And of course,this is how both van der Waerden and Emil Artin define a group in their classic presentations of algebra. Later authors of textbooks generally found this "minimalist" method of defining a group far too tedious-so they assumed the stronger axioms and called it a day. A lot of classic algebra books-Herstien's TOPICS IN ALGEBRA,famously-use this as an exercise.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:24
$begingroup$
And of course,this is how both van der Waerden and Emil Artin define a group in their classic presentations of algebra. Later authors of textbooks generally found this "minimalist" method of defining a group far too tedious-so they assumed the stronger axioms and called it a day. A lot of classic algebra books-Herstien's TOPICS IN ALGEBRA,famously-use this as an exercise.
$endgroup$
– Mathemagician1234
Sep 17 '11 at 16:24
add a comment |
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In case you don't know: Right identity and Left inverse does not imply group.
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– j.p.
Sep 17 '11 at 9:45
13
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This formulation makes the same technical error as many textbooks. The $e$ in your second axiom is not well-defined. "But obviously it's intended to be the same $e$ as in the first axiom" you reply. But the first axiom does not necessarily specify a unique element $e$. So should we interpret the second axiom as meaning "for some $e$ as in 1" or "for all $e$ as in 1"?
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– Derek Holt
Sep 17 '11 at 15:31
4
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@Derek, I think the formulation is intended to be read as "Suppose there is $ein G$ such that 1 and 2.".
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– lhf
Sep 17 '11 at 15:52
2
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@lhf: Yes, that's the formally correct way to do it, and it also removes the ambiguity.
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– Derek Holt
Sep 17 '11 at 21:29
1
$begingroup$
@ThreeFx: Take a set with two elements and define $xcdot y=x$ (possibly $x=y$), which is associative and makes both elements to right identities. If you choose a right identity, then it's also the left inverse for both elements.
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– j.p.
Jan 9 '17 at 8:02