Mixing
To prove in a Group Left identity and left inverse implies right identity and right inverse
$begingroup$
Let $G$ be a nonempty set closed under an associative product, which in addition satisfies :
A. There exists an $e$ in $G$ such that $a cdot e=a$ for all $a in G$.
B. Given $a in G$, there exists an element $y(a) in G$ such that $a cdot y(a) =e$.
Prove that $G$ must be a group under this product.
Attempt -Since Associativity is given and Closure also, also the right identity and right inverse is given .So i just have to prove left identity and left inverse.
Now as $ae=a$ post multiplying by a, $aea=aa$. Now pre multiply by a^{-1} I get hence $ea=a$. And doing same process for inverse Is this Right?
abstract-algebra group-theory proof-verification self-learning
edited Jan 11 '16 at 11:13
Martin Sleziak
44.8k9118272
asked Mar 21 '15 at 11:40
Taylor TedTaylor Ted
1,64611436
$endgroup$
add a comment |
$begingroup$
Let $G$ be a nonempty set closed under an associative product, which in addition satisfies :
A. There exists an $e$ in $G$ such that $a cdot e=a$ for all $a in G$.
B. Given $a in G$, there exists an element $y(a) in G$ such that $a cdot y(a) =e$.
Prove that $G$ must be a group under this product.
Attempt -Since Associativity is given and Closure also, also the right identity and right inverse is given .So i just have to prove left identity and left inverse.
Now as $ae=a$ post multiplying by a, $aea=aa$. Now pre multiply by a^{-1} I get hence $ea=a$. And doing same process for inverse Is this Right?
abstract-algebra group-theory proof-verification self-learning
edited Jan 11 '16 at 11:13
Martin Sleziak
44.8k9118272
asked Mar 21 '15 at 11:40
Taylor TedTaylor Ted
1,64611436
$endgroup$
$begingroup$
How are you concluding the statement after the "hence"? It looks like you're canceling, which you must prove works.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:44
$begingroup$
@MichaelBurr please check now
$endgroup$
– Taylor Ted
Mar 21 '15 at 11:45
$begingroup$
But, you're not given a left inverse. You don't know that $y(a).a=e$. You also don't know that $e.a=a$. Your proof appears circular.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:53
$begingroup$
See also: math.stackexchange.com/questions/65239/…
$endgroup$
– Martin Sleziak
Jan 11 '16 at 11:12
add a comment |
$begingroup$
Let $G$ be a nonempty set closed under an associative product, which in addition satisfies :
A. There exists an $e$ in $G$ such that $a cdot e=a$ for all $a in G$.
B. Given $a in G$, there exists an element $y(a) in G$ such that $a cdot y(a) =e$.
Prove that $G$ must be a group under this product.
Attempt -Since Associativity is given and Closure also, also the right identity and right inverse is given .So i just have to prove left identity and left inverse.
Now as $ae=a$ post multiplying by a, $aea=aa$. Now pre multiply by a^{-1} I get hence $ea=a$. And doing same process for inverse Is this Right?
abstract-algebra group-theory proof-verification self-learning
edited Jan 11 '16 at 11:13
Martin Sleziak
44.8k9118272
asked Mar 21 '15 at 11:40
Taylor TedTaylor Ted
1,64611436
$endgroup$
Let $G$ be a nonempty set closed under an associative product, which in addition satisfies :
A. There exists an $e$ in $G$ such that $a cdot e=a$ for all $a in G$.
B. Given $a in G$, there exists an element $y(a) in G$ such that $a cdot y(a) =e$.
Prove that $G$ must be a group under this product.
Attempt -Since Associativity is given and Closure also, also the right identity and right inverse is given .So i just have to prove left identity and left inverse.
Now as $ae=a$ post multiplying by a, $aea=aa$. Now pre multiply by a^{-1} I get hence $ea=a$. And doing same process for inverse Is this Right?
abstract-algebra group-theory proof-verification self-learning
abstract-algebra group-theory proof-verification self-learning
edited Jan 11 '16 at 11:13
Martin Sleziak
44.8k9118272
asked Mar 21 '15 at 11:40
Taylor TedTaylor Ted
1,64611436
edited Jan 11 '16 at 11:13
Martin Sleziak
44.8k9118272
asked Mar 21 '15 at 11:40
Taylor TedTaylor Ted
1,64611436
edited Jan 11 '16 at 11:13
Martin Sleziak
44.8k9118272
edited Jan 11 '16 at 11:13
Martin Sleziak
44.8k9118272
edited Jan 11 '16 at 11:13
Martin Sleziak
44.8k9118272
44.8k9118272
asked Mar 21 '15 at 11:40
Taylor TedTaylor Ted
1,64611436
asked Mar 21 '15 at 11:40
Taylor TedTaylor Ted
1,64611436
asked Mar 21 '15 at 11:40
Taylor TedTaylor Ted
1,64611436
1,64611436
$begingroup$
How are you concluding the statement after the "hence"? It looks like you're canceling, which you must prove works.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:44
$begingroup$
@MichaelBurr please check now
$endgroup$
– Taylor Ted
Mar 21 '15 at 11:45
$begingroup$
But, you're not given a left inverse. You don't know that $y(a).a=e$. You also don't know that $e.a=a$. Your proof appears circular.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:53
$begingroup$
See also: math.stackexchange.com/questions/65239/…
$endgroup$
– Martin Sleziak
Jan 11 '16 at 11:12
add a comment |
$begingroup$
How are you concluding the statement after the "hence"? It looks like you're canceling, which you must prove works.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:44
$begingroup$
@MichaelBurr please check now
$endgroup$
– Taylor Ted
Mar 21 '15 at 11:45
$begingroup$
But, you're not given a left inverse. You don't know that $y(a).a=e$. You also don't know that $e.a=a$. Your proof appears circular.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:53
$begingroup$
See also: math.stackexchange.com/questions/65239/…
$endgroup$
– Martin Sleziak
Jan 11 '16 at 11:12
$begingroup$
How are you concluding the statement after the "hence"? It looks like you're canceling, which you must prove works.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:44
$begingroup$
How are you concluding the statement after the "hence"? It looks like you're canceling, which you must prove works.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:44
$begingroup$
@MichaelBurr please check now
$endgroup$
– Taylor Ted
Mar 21 '15 at 11:45
$begingroup$
@MichaelBurr please check now
$endgroup$
– Taylor Ted
Mar 21 '15 at 11:45
$begingroup$
But, you're not given a left inverse. You don't know that $y(a).a=e$. You also don't know that $e.a=a$. Your proof appears circular.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:53
$begingroup$
But, you're not given a left inverse. You don't know that $y(a).a=e$. You also don't know that $e.a=a$. Your proof appears circular.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:53
$begingroup$
See also: math.stackexchange.com/questions/65239/…
$endgroup$
– Martin Sleziak
Jan 11 '16 at 11:12
$begingroup$
See also: math.stackexchange.com/questions/65239/…
$endgroup$
– Martin Sleziak
Jan 11 '16 at 11:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let, $ab=eland bc=etag {1}$
for some $b,cin G$. And, $ae=atag{2}$
From $(2)$, $$eae=eaimplies(ab)a(bc)=eaimplies ((ab)(ab))c=eaimplies ec=eatag{3}$$
Similarly, $$ae=aimplies a(bc)=aimplies (ab)c=aimplies ec=atag{4}$$
$(3)$ and $(4)$ implies, $$ea=a$$
Also from $(3)$ and $(1)$, $$(bab)(bca)=eimplies b((ab)(bc)a)=eimplies ba=e$$
answered Mar 22 '15 at 6:03
user 170039user 170039
10.4k42465
$endgroup$
$begingroup$
(1) is wrong, I think, since you pre-suppose that actually there is a left inverse. Where did you find that? It might be the case that no left inverse exists for any element. How do you prove existence?
$endgroup$
– Jason
Jul 27 '17 at 8:08
$begingroup$
Note that given $ain G$ there exists an element $y(a)in G$ such that $acdot y(a)=e$. Similar is the argument for $b$. In my answer above $y(a)=b$ and $y(b)=c$. Does it help @Jason?
$endgroup$
– user 170039
Jul 27 '17 at 13:31
$begingroup$
Can you please clarify the last assert $(bab)(bca)=e$? I fail to see how it follows from $(1)$, Thank you!
$endgroup$
– galra
Mar 3 '18 at 1:10
2
$begingroup$
@galra: See the edit. Observe that by $(3)$ we have, begin{align*}(bab)(bca)&=(be)(ea)\&=b(ec)&text{by (3)}\&=(be)c\&=bc\&=e\end{align*}And by $(1)$ we have, begin{align*}(bab)(bca)&=b(ab)(bc)a\&=b(e)(e)a\&=baend{align*} Hope it helps.
$endgroup$
– user 170039
Mar 3 '18 at 4:35
add a comment |
$begingroup$
1.
$(y(a)cdot a)cdot (y(a)cdot a) = y(a) cdot (a cdot y(a))cdot a = y(a) cdot e cdot a=(y(a)cdot e) cdot a = y(a) cdot a$.
$(y(a)cdot a)cdot ((y(a)cdot a) cdot y(y(a) cdot a)) = (y(a) cdot a) cdot y(y(a) cdot a)$.
$y(a)cdot a = e$
2.
$ecdot a = (a cdot y(a))cdot a=acdot(y(a)cdot a)=acdot e=a$.
answered Jan 9 at 2:57
tchappy hatchappy ha
647311
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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oldest
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$begingroup$
Let, $ab=eland bc=etag {1}$
for some $b,cin G$. And, $ae=atag{2}$
From $(2)$, $$eae=eaimplies(ab)a(bc)=eaimplies ((ab)(ab))c=eaimplies ec=eatag{3}$$
Similarly, $$ae=aimplies a(bc)=aimplies (ab)c=aimplies ec=atag{4}$$
$(3)$ and $(4)$ implies, $$ea=a$$
Also from $(3)$ and $(1)$, $$(bab)(bca)=eimplies b((ab)(bc)a)=eimplies ba=e$$
answered Mar 22 '15 at 6:03
user 170039user 170039
10.4k42465
$endgroup$
$begingroup$
(1) is wrong, I think, since you pre-suppose that actually there is a left inverse. Where did you find that? It might be the case that no left inverse exists for any element. How do you prove existence?
$endgroup$
– Jason
Jul 27 '17 at 8:08
$begingroup$
Note that given $ain G$ there exists an element $y(a)in G$ such that $acdot y(a)=e$. Similar is the argument for $b$. In my answer above $y(a)=b$ and $y(b)=c$. Does it help @Jason?
$endgroup$
– user 170039
Jul 27 '17 at 13:31
$begingroup$
Can you please clarify the last assert $(bab)(bca)=e$? I fail to see how it follows from $(1)$, Thank you!
$endgroup$
– galra
Mar 3 '18 at 1:10
2
$begingroup$
@galra: See the edit. Observe that by $(3)$ we have, begin{align*}(bab)(bca)&=(be)(ea)\&=b(ec)&text{by (3)}\&=(be)c\&=bc\&=e\end{align*}And by $(1)$ we have, begin{align*}(bab)(bca)&=b(ab)(bc)a\&=b(e)(e)a\&=baend{align*} Hope it helps.
$endgroup$
– user 170039
Mar 3 '18 at 4:35
add a comment |
$begingroup$
Let, $ab=eland bc=etag {1}$
for some $b,cin G$. And, $ae=atag{2}$
From $(2)$, $$eae=eaimplies(ab)a(bc)=eaimplies ((ab)(ab))c=eaimplies ec=eatag{3}$$
Similarly, $$ae=aimplies a(bc)=aimplies (ab)c=aimplies ec=atag{4}$$
$(3)$ and $(4)$ implies, $$ea=a$$
Also from $(3)$ and $(1)$, $$(bab)(bca)=eimplies b((ab)(bc)a)=eimplies ba=e$$
answered Mar 22 '15 at 6:03
user 170039user 170039
10.4k42465
$endgroup$
$begingroup$
(1) is wrong, I think, since you pre-suppose that actually there is a left inverse. Where did you find that? It might be the case that no left inverse exists for any element. How do you prove existence?
$endgroup$
– Jason
Jul 27 '17 at 8:08
$begingroup$
Note that given $ain G$ there exists an element $y(a)in G$ such that $acdot y(a)=e$. Similar is the argument for $b$. In my answer above $y(a)=b$ and $y(b)=c$. Does it help @Jason?
$endgroup$
– user 170039
Jul 27 '17 at 13:31
$begingroup$
Can you please clarify the last assert $(bab)(bca)=e$? I fail to see how it follows from $(1)$, Thank you!
$endgroup$
– galra
Mar 3 '18 at 1:10
2
$begingroup$
@galra: See the edit. Observe that by $(3)$ we have, begin{align*}(bab)(bca)&=(be)(ea)\&=b(ec)&text{by (3)}\&=(be)c\&=bc\&=e\end{align*}And by $(1)$ we have, begin{align*}(bab)(bca)&=b(ab)(bc)a\&=b(e)(e)a\&=baend{align*} Hope it helps.
$endgroup$
– user 170039
Mar 3 '18 at 4:35
add a comment |
$begingroup$
Let, $ab=eland bc=etag {1}$
for some $b,cin G$. And, $ae=atag{2}$
From $(2)$, $$eae=eaimplies(ab)a(bc)=eaimplies ((ab)(ab))c=eaimplies ec=eatag{3}$$
Similarly, $$ae=aimplies a(bc)=aimplies (ab)c=aimplies ec=atag{4}$$
$(3)$ and $(4)$ implies, $$ea=a$$
Also from $(3)$ and $(1)$, $$(bab)(bca)=eimplies b((ab)(bc)a)=eimplies ba=e$$
answered Mar 22 '15 at 6:03
user 170039user 170039
10.4k42465
$endgroup$
Let, $ab=eland bc=etag {1}$
for some $b,cin G$. And, $ae=atag{2}$
From $(2)$, $$eae=eaimplies(ab)a(bc)=eaimplies ((ab)(ab))c=eaimplies ec=eatag{3}$$
Similarly, $$ae=aimplies a(bc)=aimplies (ab)c=aimplies ec=atag{4}$$
$(3)$ and $(4)$ implies, $$ea=a$$
Also from $(3)$ and $(1)$, $$(bab)(bca)=eimplies b((ab)(bc)a)=eimplies ba=e$$
answered Mar 22 '15 at 6:03
user 170039user 170039
10.4k42465
edited Mar 3 '18 at 4:31
answered Mar 22 '15 at 6:03
user 170039user 170039
10.4k42465
answered Mar 22 '15 at 6:03
user 170039user 170039
10.4k42465
answered Mar 22 '15 at 6:03
user 170039user 170039
10.4k42465
10.4k42465
$begingroup$
(1) is wrong, I think, since you pre-suppose that actually there is a left inverse. Where did you find that? It might be the case that no left inverse exists for any element. How do you prove existence?
$endgroup$
– Jason
Jul 27 '17 at 8:08
$begingroup$
Note that given $ain G$ there exists an element $y(a)in G$ such that $acdot y(a)=e$. Similar is the argument for $b$. In my answer above $y(a)=b$ and $y(b)=c$. Does it help @Jason?
$endgroup$
– user 170039
Jul 27 '17 at 13:31
$begingroup$
Can you please clarify the last assert $(bab)(bca)=e$? I fail to see how it follows from $(1)$, Thank you!
$endgroup$
– galra
Mar 3 '18 at 1:10
2
$begingroup$
@galra: See the edit. Observe that by $(3)$ we have, begin{align*}(bab)(bca)&=(be)(ea)\&=b(ec)&text{by (3)}\&=(be)c\&=bc\&=e\end{align*}And by $(1)$ we have, begin{align*}(bab)(bca)&=b(ab)(bc)a\&=b(e)(e)a\&=baend{align*} Hope it helps.
$endgroup$
– user 170039
Mar 3 '18 at 4:35
add a comment |
$begingroup$
(1) is wrong, I think, since you pre-suppose that actually there is a left inverse. Where did you find that? It might be the case that no left inverse exists for any element. How do you prove existence?
$endgroup$
– Jason
Jul 27 '17 at 8:08
$begingroup$
Note that given $ain G$ there exists an element $y(a)in G$ such that $acdot y(a)=e$. Similar is the argument for $b$. In my answer above $y(a)=b$ and $y(b)=c$. Does it help @Jason?
$endgroup$
– user 170039
Jul 27 '17 at 13:31
$begingroup$
Can you please clarify the last assert $(bab)(bca)=e$? I fail to see how it follows from $(1)$, Thank you!
$endgroup$
– galra
Mar 3 '18 at 1:10
2
$begingroup$
@galra: See the edit. Observe that by $(3)$ we have, begin{align*}(bab)(bca)&=(be)(ea)\&=b(ec)&text{by (3)}\&=(be)c\&=bc\&=e\end{align*}And by $(1)$ we have, begin{align*}(bab)(bca)&=b(ab)(bc)a\&=b(e)(e)a\&=baend{align*} Hope it helps.
$endgroup$
– user 170039
Mar 3 '18 at 4:35
$begingroup$
(1) is wrong, I think, since you pre-suppose that actually there is a left inverse. Where did you find that? It might be the case that no left inverse exists for any element. How do you prove existence?
$endgroup$
– Jason
Jul 27 '17 at 8:08
$begingroup$
(1) is wrong, I think, since you pre-suppose that actually there is a left inverse. Where did you find that? It might be the case that no left inverse exists for any element. How do you prove existence?
$endgroup$
– Jason
Jul 27 '17 at 8:08
$begingroup$
Note that given $ain G$ there exists an element $y(a)in G$ such that $acdot y(a)=e$. Similar is the argument for $b$. In my answer above $y(a)=b$ and $y(b)=c$. Does it help @Jason?
$endgroup$
– user 170039
Jul 27 '17 at 13:31
$begingroup$
Note that given $ain G$ there exists an element $y(a)in G$ such that $acdot y(a)=e$. Similar is the argument for $b$. In my answer above $y(a)=b$ and $y(b)=c$. Does it help @Jason?
$endgroup$
– user 170039
Jul 27 '17 at 13:31
$begingroup$
Can you please clarify the last assert $(bab)(bca)=e$? I fail to see how it follows from $(1)$, Thank you!
$endgroup$
– galra
Mar 3 '18 at 1:10
$begingroup$
Can you please clarify the last assert $(bab)(bca)=e$? I fail to see how it follows from $(1)$, Thank you!
$endgroup$
– galra
Mar 3 '18 at 1:10
2
2
$begingroup$
@galra: See the edit. Observe that by $(3)$ we have, begin{align*}(bab)(bca)&=(be)(ea)\&=b(ec)&text{by (3)}\&=(be)c\&=bc\&=e\end{align*}And by $(1)$ we have, begin{align*}(bab)(bca)&=b(ab)(bc)a\&=b(e)(e)a\&=baend{align*} Hope it helps.
$endgroup$
– user 170039
Mar 3 '18 at 4:35
$begingroup$
@galra: See the edit. Observe that by $(3)$ we have, begin{align*}(bab)(bca)&=(be)(ea)\&=b(ec)&text{by (3)}\&=(be)c\&=bc\&=e\end{align*}And by $(1)$ we have, begin{align*}(bab)(bca)&=b(ab)(bc)a\&=b(e)(e)a\&=baend{align*} Hope it helps.
$endgroup$
– user 170039
Mar 3 '18 at 4:35
add a comment |
$begingroup$
1.
$(y(a)cdot a)cdot (y(a)cdot a) = y(a) cdot (a cdot y(a))cdot a = y(a) cdot e cdot a=(y(a)cdot e) cdot a = y(a) cdot a$.
$(y(a)cdot a)cdot ((y(a)cdot a) cdot y(y(a) cdot a)) = (y(a) cdot a) cdot y(y(a) cdot a)$.
$y(a)cdot a = e$
2.
$ecdot a = (a cdot y(a))cdot a=acdot(y(a)cdot a)=acdot e=a$.
answered Jan 9 at 2:57
tchappy hatchappy ha
647311
$endgroup$
add a comment |
$begingroup$
1.
$(y(a)cdot a)cdot (y(a)cdot a) = y(a) cdot (a cdot y(a))cdot a = y(a) cdot e cdot a=(y(a)cdot e) cdot a = y(a) cdot a$.
$(y(a)cdot a)cdot ((y(a)cdot a) cdot y(y(a) cdot a)) = (y(a) cdot a) cdot y(y(a) cdot a)$.
$y(a)cdot a = e$
2.
$ecdot a = (a cdot y(a))cdot a=acdot(y(a)cdot a)=acdot e=a$.
answered Jan 9 at 2:57
tchappy hatchappy ha
647311
$endgroup$
add a comment |
$begingroup$
1.
$(y(a)cdot a)cdot (y(a)cdot a) = y(a) cdot (a cdot y(a))cdot a = y(a) cdot e cdot a=(y(a)cdot e) cdot a = y(a) cdot a$.
$(y(a)cdot a)cdot ((y(a)cdot a) cdot y(y(a) cdot a)) = (y(a) cdot a) cdot y(y(a) cdot a)$.
$y(a)cdot a = e$
2.
$ecdot a = (a cdot y(a))cdot a=acdot(y(a)cdot a)=acdot e=a$.
answered Jan 9 at 2:57
tchappy hatchappy ha
647311
$endgroup$
1.
$(y(a)cdot a)cdot (y(a)cdot a) = y(a) cdot (a cdot y(a))cdot a = y(a) cdot e cdot a=(y(a)cdot e) cdot a = y(a) cdot a$.
$(y(a)cdot a)cdot ((y(a)cdot a) cdot y(y(a) cdot a)) = (y(a) cdot a) cdot y(y(a) cdot a)$.
$y(a)cdot a = e$
2.
$ecdot a = (a cdot y(a))cdot a=acdot(y(a)cdot a)=acdot e=a$.
answered Jan 9 at 2:57
tchappy hatchappy ha
647311
answered Jan 9 at 2:57
tchappy hatchappy ha
647311
answered Jan 9 at 2:57
tchappy hatchappy ha
647311
answered Jan 9 at 2:57
tchappy hatchappy ha
647311
647311
add a comment |
add a comment |
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$begingroup$
How are you concluding the statement after the "hence"? It looks like you're canceling, which you must prove works.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:44
$begingroup$
@MichaelBurr please check now
$endgroup$
– Taylor Ted
Mar 21 '15 at 11:45
$begingroup$
But, you're not given a left inverse. You don't know that $y(a).a=e$. You also don't know that $e.a=a$. Your proof appears circular.
$endgroup$
– Michael Burr
Mar 21 '15 at 11:53
$begingroup$
See also: math.stackexchange.com/questions/65239/…
$endgroup$
– Martin Sleziak
Jan 11 '16 at 11:12