Mixing
Set of all inner automorphisms is a normal subgroup
$begingroup$
In order to prove this, I first proved that the set of all automorphisms from a group $G$ to $G$ form a group under composition:
The identity homorphism is an automorphism because sends $x$ from $G$ to $x$ from $G$:
$phi_e(x) = x in G$
The inverse of an homomorphism exists:
well...
The associativity works because composition of functions is always associative.
Closure property:
$$phicircgamma(a+b) = phi(gamma(a+b)) = phi(gamma(a)+gamma(b)) = phi(gamma(a)) + phi(gamma(a)) = phicircgamma(a)+phicircgamma(b)$$
So, an Inner automorphism is defined as a function $f$ such that
$f(x) = a^{-1}xa$
for a fixed element $a$ from $G$.
I'm supposed to prove that the set of all these automorphisms form a normal subgroup of $G$, that is:
$$gfg^{-1}in N$$ for all $g$
where $N$ is the set of all inner automorphisms of $G$, and $g$ is an automorphism of $G$.
UPDATE:
Ok, so what I learned from this is that
we have a group $Aut(G)$ made of all the automorphisms of $G$, and we
want to show that the subgroup of $Aut(G)$ made of all the inner
automorphisms, is normal. That is, given an inner automorphism
$phi_a(x) = a^{-1}xa$ for a fixed $a$ in $G$ and $xin G$.
So, we need to show that, given $g$ as an automorphism from $Aut(G)$
and $phi_a$ an inner automorphism from the subgroup $N$ of inner
automorphisms, we must have:
$$g^{-1}phi_ag in N$$
for all $g$
Am I rigth?
abstract-algebra group-theory finite-groups
asked Aug 22 '15 at 21:58
Guerlando OCsGuerlando OCs
13621654
$endgroup$
|
show 1 more comment
$begingroup$
In order to prove this, I first proved that the set of all automorphisms from a group $G$ to $G$ form a group under composition:
The identity homorphism is an automorphism because sends $x$ from $G$ to $x$ from $G$:
$phi_e(x) = x in G$
The inverse of an homomorphism exists:
well...
The associativity works because composition of functions is always associative.
Closure property:
$$phicircgamma(a+b) = phi(gamma(a+b)) = phi(gamma(a)+gamma(b)) = phi(gamma(a)) + phi(gamma(a)) = phicircgamma(a)+phicircgamma(b)$$
So, an Inner automorphism is defined as a function $f$ such that
$f(x) = a^{-1}xa$
for a fixed element $a$ from $G$.
I'm supposed to prove that the set of all these automorphisms form a normal subgroup of $G$, that is:
$$gfg^{-1}in N$$ for all $g$
where $N$ is the set of all inner automorphisms of $G$, and $g$ is an automorphism of $G$.
UPDATE:
Ok, so what I learned from this is that
we have a group $Aut(G)$ made of all the automorphisms of $G$, and we
want to show that the subgroup of $Aut(G)$ made of all the inner
automorphisms, is normal. That is, given an inner automorphism
$phi_a(x) = a^{-1}xa$ for a fixed $a$ in $G$ and $xin G$.
So, we need to show that, given $g$ as an automorphism from $Aut(G)$
and $phi_a$ an inner automorphism from the subgroup $N$ of inner
automorphisms, we must have:
$$g^{-1}phi_ag in N$$
for all $g$
Am I rigth?
abstract-algebra group-theory finite-groups
asked Aug 22 '15 at 21:58
Guerlando OCsGuerlando OCs
13621654
$endgroup$
3
$begingroup$
For clarity, instead of $f$, let $varphi_a$ denote the inner automorphism $varphi_a(x)=a^{-1}xa$. Then show $gvarphi_a g^{-1}=varphi_{g(a)}$, hence $gvarphi_a g^{-1}$ is inner.
$endgroup$
– Ben West
Aug 22 '15 at 22:04
4
$begingroup$
It's a normal subgroup of $operatorname{Aut}G$, not of $G$.
$endgroup$
– Bernard
Aug 22 '15 at 22:11
2
$begingroup$
Your title is deceptive. The question is about inner automorphisms from a group to itself, not to another group. In fact, one only uses the terminology "automorphisms" when talking about isomorphisms from an object to itself.
$endgroup$
– Lee Mosher
Aug 22 '15 at 22:33
$begingroup$
$g(a^{-1}xa)g^{-1} = ga^{-1}xag^{-1} = (ag^{-1})^{-1}x(ga^{-1})^1$ but I don't think this help. Can I argue that $ag^{-1}$ is in $G$?
$endgroup$
– Guerlando OCs
Aug 22 '15 at 23:43
1
$begingroup$
@GuerlandoOCs No, $g$ is an automorphism of $G$, not an element of $G$, so $$gvarphi_ag^{-1}(x)=gvarphi_a(g^{-1}(x))=g(a^{-1}g^{-1}(x)a)=g(a^{-1})g(g^{-1}(x))g(a)=g(a)^{-1}xg(a)=varphi_{g(a)}(x).$$
$endgroup$
– Ben West
Aug 23 '15 at 0:03
|
show 1 more comment
$begingroup$
In order to prove this, I first proved that the set of all automorphisms from a group $G$ to $G$ form a group under composition:
The identity homorphism is an automorphism because sends $x$ from $G$ to $x$ from $G$:
$phi_e(x) = x in G$
The inverse of an homomorphism exists:
well...
The associativity works because composition of functions is always associative.
Closure property:
$$phicircgamma(a+b) = phi(gamma(a+b)) = phi(gamma(a)+gamma(b)) = phi(gamma(a)) + phi(gamma(a)) = phicircgamma(a)+phicircgamma(b)$$
So, an Inner automorphism is defined as a function $f$ such that
$f(x) = a^{-1}xa$
for a fixed element $a$ from $G$.
I'm supposed to prove that the set of all these automorphisms form a normal subgroup of $G$, that is:
$$gfg^{-1}in N$$ for all $g$
where $N$ is the set of all inner automorphisms of $G$, and $g$ is an automorphism of $G$.
UPDATE:
Ok, so what I learned from this is that
we have a group $Aut(G)$ made of all the automorphisms of $G$, and we
want to show that the subgroup of $Aut(G)$ made of all the inner
automorphisms, is normal. That is, given an inner automorphism
$phi_a(x) = a^{-1}xa$ for a fixed $a$ in $G$ and $xin G$.
So, we need to show that, given $g$ as an automorphism from $Aut(G)$
and $phi_a$ an inner automorphism from the subgroup $N$ of inner
automorphisms, we must have:
$$g^{-1}phi_ag in N$$
for all $g$
Am I rigth?
abstract-algebra group-theory finite-groups
asked Aug 22 '15 at 21:58
Guerlando OCsGuerlando OCs
13621654
$endgroup$
In order to prove this, I first proved that the set of all automorphisms from a group $G$ to $G$ form a group under composition:
The identity homorphism is an automorphism because sends $x$ from $G$ to $x$ from $G$:
$phi_e(x) = x in G$
The inverse of an homomorphism exists:
well...
The associativity works because composition of functions is always associative.
Closure property:
$$phicircgamma(a+b) = phi(gamma(a+b)) = phi(gamma(a)+gamma(b)) = phi(gamma(a)) + phi(gamma(a)) = phicircgamma(a)+phicircgamma(b)$$
So, an Inner automorphism is defined as a function $f$ such that
$f(x) = a^{-1}xa$
for a fixed element $a$ from $G$.
I'm supposed to prove that the set of all these automorphisms form a normal subgroup of $G$, that is:
$$gfg^{-1}in N$$ for all $g$
where $N$ is the set of all inner automorphisms of $G$, and $g$ is an automorphism of $G$.
UPDATE:
Ok, so what I learned from this is that
we have a group $Aut(G)$ made of all the automorphisms of $G$, and we
want to show that the subgroup of $Aut(G)$ made of all the inner
automorphisms, is normal. That is, given an inner automorphism
$phi_a(x) = a^{-1}xa$ for a fixed $a$ in $G$ and $xin G$.
So, we need to show that, given $g$ as an automorphism from $Aut(G)$
and $phi_a$ an inner automorphism from the subgroup $N$ of inner
automorphisms, we must have:
$$g^{-1}phi_ag in N$$
for all $g$
Am I rigth?
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked Aug 22 '15 at 21:58
Guerlando OCsGuerlando OCs
13621654
asked Aug 22 '15 at 21:58
Guerlando OCsGuerlando OCs
13621654
edited Aug 23 '15 at 0:17
Guerlando OCs
asked Aug 22 '15 at 21:58
Guerlando OCsGuerlando OCs
13621654
asked Aug 22 '15 at 21:58
Guerlando OCsGuerlando OCs
13621654
asked Aug 22 '15 at 21:58
Guerlando OCsGuerlando OCs
13621654
13621654
3
$begingroup$
For clarity, instead of $f$, let $varphi_a$ denote the inner automorphism $varphi_a(x)=a^{-1}xa$. Then show $gvarphi_a g^{-1}=varphi_{g(a)}$, hence $gvarphi_a g^{-1}$ is inner.
$endgroup$
– Ben West
Aug 22 '15 at 22:04
4
$begingroup$
It's a normal subgroup of $operatorname{Aut}G$, not of $G$.
$endgroup$
– Bernard
Aug 22 '15 at 22:11
2
$begingroup$
Your title is deceptive. The question is about inner automorphisms from a group to itself, not to another group. In fact, one only uses the terminology "automorphisms" when talking about isomorphisms from an object to itself.
$endgroup$
– Lee Mosher
Aug 22 '15 at 22:33
$begingroup$
$g(a^{-1}xa)g^{-1} = ga^{-1}xag^{-1} = (ag^{-1})^{-1}x(ga^{-1})^1$ but I don't think this help. Can I argue that $ag^{-1}$ is in $G$?
$endgroup$
– Guerlando OCs
Aug 22 '15 at 23:43
1
$begingroup$
@GuerlandoOCs No, $g$ is an automorphism of $G$, not an element of $G$, so $$gvarphi_ag^{-1}(x)=gvarphi_a(g^{-1}(x))=g(a^{-1}g^{-1}(x)a)=g(a^{-1})g(g^{-1}(x))g(a)=g(a)^{-1}xg(a)=varphi_{g(a)}(x).$$
$endgroup$
– Ben West
Aug 23 '15 at 0:03
|
show 1 more comment
3
$begingroup$
For clarity, instead of $f$, let $varphi_a$ denote the inner automorphism $varphi_a(x)=a^{-1}xa$. Then show $gvarphi_a g^{-1}=varphi_{g(a)}$, hence $gvarphi_a g^{-1}$ is inner.
$endgroup$
– Ben West
Aug 22 '15 at 22:04
4
$begingroup$
It's a normal subgroup of $operatorname{Aut}G$, not of $G$.
$endgroup$
– Bernard
Aug 22 '15 at 22:11
2
$begingroup$
Your title is deceptive. The question is about inner automorphisms from a group to itself, not to another group. In fact, one only uses the terminology "automorphisms" when talking about isomorphisms from an object to itself.
$endgroup$
– Lee Mosher
Aug 22 '15 at 22:33
$begingroup$
$g(a^{-1}xa)g^{-1} = ga^{-1}xag^{-1} = (ag^{-1})^{-1}x(ga^{-1})^1$ but I don't think this help. Can I argue that $ag^{-1}$ is in $G$?
$endgroup$
– Guerlando OCs
Aug 22 '15 at 23:43
1
$begingroup$
@GuerlandoOCs No, $g$ is an automorphism of $G$, not an element of $G$, so $$gvarphi_ag^{-1}(x)=gvarphi_a(g^{-1}(x))=g(a^{-1}g^{-1}(x)a)=g(a^{-1})g(g^{-1}(x))g(a)=g(a)^{-1}xg(a)=varphi_{g(a)}(x).$$
$endgroup$
– Ben West
Aug 23 '15 at 0:03
3
3
$begingroup$
For clarity, instead of $f$, let $varphi_a$ denote the inner automorphism $varphi_a(x)=a^{-1}xa$. Then show $gvarphi_a g^{-1}=varphi_{g(a)}$, hence $gvarphi_a g^{-1}$ is inner.
$endgroup$
– Ben West
Aug 22 '15 at 22:04
$begingroup$
For clarity, instead of $f$, let $varphi_a$ denote the inner automorphism $varphi_a(x)=a^{-1}xa$. Then show $gvarphi_a g^{-1}=varphi_{g(a)}$, hence $gvarphi_a g^{-1}$ is inner.
$endgroup$
– Ben West
Aug 22 '15 at 22:04
4
4
$begingroup$
It's a normal subgroup of $operatorname{Aut}G$, not of $G$.
$endgroup$
– Bernard
Aug 22 '15 at 22:11
$begingroup$
It's a normal subgroup of $operatorname{Aut}G$, not of $G$.
$endgroup$
– Bernard
Aug 22 '15 at 22:11
2
2
$begingroup$
Your title is deceptive. The question is about inner automorphisms from a group to itself, not to another group. In fact, one only uses the terminology "automorphisms" when talking about isomorphisms from an object to itself.
$endgroup$
– Lee Mosher
Aug 22 '15 at 22:33
$begingroup$
Your title is deceptive. The question is about inner automorphisms from a group to itself, not to another group. In fact, one only uses the terminology "automorphisms" when talking about isomorphisms from an object to itself.
$endgroup$
– Lee Mosher
Aug 22 '15 at 22:33
$begingroup$
$g(a^{-1}xa)g^{-1} = ga^{-1}xag^{-1} = (ag^{-1})^{-1}x(ga^{-1})^1$ but I don't think this help. Can I argue that $ag^{-1}$ is in $G$?
$endgroup$
– Guerlando OCs
Aug 22 '15 at 23:43
$begingroup$
$g(a^{-1}xa)g^{-1} = ga^{-1}xag^{-1} = (ag^{-1})^{-1}x(ga^{-1})^1$ but I don't think this help. Can I argue that $ag^{-1}$ is in $G$?
$endgroup$
– Guerlando OCs
Aug 22 '15 at 23:43
1
1
$begingroup$
@GuerlandoOCs No, $g$ is an automorphism of $G$, not an element of $G$, so $$gvarphi_ag^{-1}(x)=gvarphi_a(g^{-1}(x))=g(a^{-1}g^{-1}(x)a)=g(a^{-1})g(g^{-1}(x))g(a)=g(a)^{-1}xg(a)=varphi_{g(a)}(x).$$
$endgroup$
– Ben West
Aug 23 '15 at 0:03
$begingroup$
@GuerlandoOCs No, $g$ is an automorphism of $G$, not an element of $G$, so $$gvarphi_ag^{-1}(x)=gvarphi_a(g^{-1}(x))=g(a^{-1}g^{-1}(x)a)=g(a^{-1})g(g^{-1}(x))g(a)=g(a)^{-1}xg(a)=varphi_{g(a)}(x).$$
$endgroup$
– Ben West
Aug 23 '15 at 0:03
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
To prove "$mathrm{Inn}(G)$ is a normal subgroup of $mathrm{Aut}(G)$", you need to prove "for each $phi_ain mathrm{Inn}(G)$ and $theta in mathrm{Aut}(G)$, we have $theta circ phi_a circ theta^{-1} in mathrm{Inn}(G)$". To verify this, simply evaluate $(theta circ phi_a circ theta^{-1})(x)$ and use some properties that $theta$ and $phi_a$ are known to have.
answered Jun 27 '18 at 14:05
Clay ThomasClay Thomas
1406
$endgroup$
add a comment |
$begingroup$
First, you used additive notation for the group operation, but we needn't suppose that the group $G$ is abelian, the result holds in the general case.
Second, group automorphisms $Aut(G)$ is a different group from $G$ itself (so, inner automorphisms do not form a subgroup in $G$...).
The set/class of all inner automorphisms $Inn(G)$ form a normal subgroup in $Aut(G)$, not in $G$ itself!
Conjugation operation, you called it $f(x)$, more precisely $f_a(x)$ or $x^a$, is a group homomorphism $Grightarrow Aut(G)$ (not necessarily isomorphism or even mono- or epi-).
And the whole image of a group homomorphism is always a normal subgroup in the codomain (the proof is a bit verbose).
In this case the whole image of the conjugation operation is exactly the set of all inner automorphisms (forming a subgroup in $Aut(G)$).
$$ f: Grightarrow Aut(G) $$
$$ f_a(x) := a^{-1}xa $$
$$ f_G = Inn(G) $$
$$ Inn(G) triangleleft Aut(G) $$
So we can indeed always speak about outer automorphisms $Out(G)$ of a group $G$, which is possibly trivial, because $Inn(G)$ exhaust the whole $Aut(G)$ (all automorphisms are inner), or, conversely, equal (isomorphic) to $Aut(G)$, because $Inn(G)$ is trivial (in case of abelian groups).
answered Aug 22 '15 at 23:55
Andrew MiloradovskyAndrew Miloradovsky
37418
$endgroup$
$begingroup$
PS: Actually, we should use pairing (carrier, oepration(s)) to denote groups without ambiguity, but we assume, as always, that the carrier and the algebraic structure are denoted by the same symbol.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:03
$begingroup$
Also, $Inn(G)$ is isomorphic to $G/Z(G)$, quotient group to the kernel of the conjugation homomorphism. Maybe you meant this.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:08
$begingroup$
Also, it is considerably a bad idea to denote $Inn(G)$ by $N$, because just N suggests that it is a subgroup of G, but it isn't.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:24
$begingroup$
And also, to prove this result you don't have to do it directly by definition of normality (that it contains also all the conjugate elements). Because you must have strict distinction which elements are from $G$ and which are from $Aut(G)$. For example, your $g$, performing conjugation in $Aut(G)$ is also an automorphism (not necessarily inner), but not and element of $G$.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:26
$begingroup$
You write "And the whole image of a group homomorphism is always a normal subgroup in the codomain". This is false. Take, as a counterexample, any non-normal subgroup $H le G$ and consider the inclusion map $H to G$ given simply by $i(h) = h$.
$endgroup$
– Clay Thomas
Jun 27 '18 at 13:59
add a comment |
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2 Answers
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2 Answers
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active
oldest
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$begingroup$
To prove "$mathrm{Inn}(G)$ is a normal subgroup of $mathrm{Aut}(G)$", you need to prove "for each $phi_ain mathrm{Inn}(G)$ and $theta in mathrm{Aut}(G)$, we have $theta circ phi_a circ theta^{-1} in mathrm{Inn}(G)$". To verify this, simply evaluate $(theta circ phi_a circ theta^{-1})(x)$ and use some properties that $theta$ and $phi_a$ are known to have.
answered Jun 27 '18 at 14:05
Clay ThomasClay Thomas
1406
$endgroup$
add a comment |
$begingroup$
To prove "$mathrm{Inn}(G)$ is a normal subgroup of $mathrm{Aut}(G)$", you need to prove "for each $phi_ain mathrm{Inn}(G)$ and $theta in mathrm{Aut}(G)$, we have $theta circ phi_a circ theta^{-1} in mathrm{Inn}(G)$". To verify this, simply evaluate $(theta circ phi_a circ theta^{-1})(x)$ and use some properties that $theta$ and $phi_a$ are known to have.
answered Jun 27 '18 at 14:05
Clay ThomasClay Thomas
1406
$endgroup$
add a comment |
$begingroup$
To prove "$mathrm{Inn}(G)$ is a normal subgroup of $mathrm{Aut}(G)$", you need to prove "for each $phi_ain mathrm{Inn}(G)$ and $theta in mathrm{Aut}(G)$, we have $theta circ phi_a circ theta^{-1} in mathrm{Inn}(G)$". To verify this, simply evaluate $(theta circ phi_a circ theta^{-1})(x)$ and use some properties that $theta$ and $phi_a$ are known to have.
answered Jun 27 '18 at 14:05
Clay ThomasClay Thomas
1406
$endgroup$
To prove "$mathrm{Inn}(G)$ is a normal subgroup of $mathrm{Aut}(G)$", you need to prove "for each $phi_ain mathrm{Inn}(G)$ and $theta in mathrm{Aut}(G)$, we have $theta circ phi_a circ theta^{-1} in mathrm{Inn}(G)$". To verify this, simply evaluate $(theta circ phi_a circ theta^{-1})(x)$ and use some properties that $theta$ and $phi_a$ are known to have.
answered Jun 27 '18 at 14:05
Clay ThomasClay Thomas
1406
answered Jun 27 '18 at 14:05
Clay ThomasClay Thomas
1406
answered Jun 27 '18 at 14:05
Clay ThomasClay Thomas
1406
answered Jun 27 '18 at 14:05
Clay ThomasClay Thomas
1406
1406
add a comment |
add a comment |
$begingroup$
First, you used additive notation for the group operation, but we needn't suppose that the group $G$ is abelian, the result holds in the general case.
Second, group automorphisms $Aut(G)$ is a different group from $G$ itself (so, inner automorphisms do not form a subgroup in $G$...).
The set/class of all inner automorphisms $Inn(G)$ form a normal subgroup in $Aut(G)$, not in $G$ itself!
Conjugation operation, you called it $f(x)$, more precisely $f_a(x)$ or $x^a$, is a group homomorphism $Grightarrow Aut(G)$ (not necessarily isomorphism or even mono- or epi-).
And the whole image of a group homomorphism is always a normal subgroup in the codomain (the proof is a bit verbose).
In this case the whole image of the conjugation operation is exactly the set of all inner automorphisms (forming a subgroup in $Aut(G)$).
$$ f: Grightarrow Aut(G) $$
$$ f_a(x) := a^{-1}xa $$
$$ f_G = Inn(G) $$
$$ Inn(G) triangleleft Aut(G) $$
So we can indeed always speak about outer automorphisms $Out(G)$ of a group $G$, which is possibly trivial, because $Inn(G)$ exhaust the whole $Aut(G)$ (all automorphisms are inner), or, conversely, equal (isomorphic) to $Aut(G)$, because $Inn(G)$ is trivial (in case of abelian groups).
answered Aug 22 '15 at 23:55
Andrew MiloradovskyAndrew Miloradovsky
37418
$endgroup$
$begingroup$
PS: Actually, we should use pairing (carrier, oepration(s)) to denote groups without ambiguity, but we assume, as always, that the carrier and the algebraic structure are denoted by the same symbol.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:03
$begingroup$
Also, $Inn(G)$ is isomorphic to $G/Z(G)$, quotient group to the kernel of the conjugation homomorphism. Maybe you meant this.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:08
$begingroup$
Also, it is considerably a bad idea to denote $Inn(G)$ by $N$, because just N suggests that it is a subgroup of G, but it isn't.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:24
$begingroup$
And also, to prove this result you don't have to do it directly by definition of normality (that it contains also all the conjugate elements). Because you must have strict distinction which elements are from $G$ and which are from $Aut(G)$. For example, your $g$, performing conjugation in $Aut(G)$ is also an automorphism (not necessarily inner), but not and element of $G$.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:26
$begingroup$
You write "And the whole image of a group homomorphism is always a normal subgroup in the codomain". This is false. Take, as a counterexample, any non-normal subgroup $H le G$ and consider the inclusion map $H to G$ given simply by $i(h) = h$.
$endgroup$
– Clay Thomas
Jun 27 '18 at 13:59
add a comment |
$begingroup$
First, you used additive notation for the group operation, but we needn't suppose that the group $G$ is abelian, the result holds in the general case.
Second, group automorphisms $Aut(G)$ is a different group from $G$ itself (so, inner automorphisms do not form a subgroup in $G$...).
The set/class of all inner automorphisms $Inn(G)$ form a normal subgroup in $Aut(G)$, not in $G$ itself!
Conjugation operation, you called it $f(x)$, more precisely $f_a(x)$ or $x^a$, is a group homomorphism $Grightarrow Aut(G)$ (not necessarily isomorphism or even mono- or epi-).
And the whole image of a group homomorphism is always a normal subgroup in the codomain (the proof is a bit verbose).
In this case the whole image of the conjugation operation is exactly the set of all inner automorphisms (forming a subgroup in $Aut(G)$).
$$ f: Grightarrow Aut(G) $$
$$ f_a(x) := a^{-1}xa $$
$$ f_G = Inn(G) $$
$$ Inn(G) triangleleft Aut(G) $$
So we can indeed always speak about outer automorphisms $Out(G)$ of a group $G$, which is possibly trivial, because $Inn(G)$ exhaust the whole $Aut(G)$ (all automorphisms are inner), or, conversely, equal (isomorphic) to $Aut(G)$, because $Inn(G)$ is trivial (in case of abelian groups).
answered Aug 22 '15 at 23:55
Andrew MiloradovskyAndrew Miloradovsky
37418
$endgroup$
$begingroup$
PS: Actually, we should use pairing (carrier, oepration(s)) to denote groups without ambiguity, but we assume, as always, that the carrier and the algebraic structure are denoted by the same symbol.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:03
$begingroup$
Also, $Inn(G)$ is isomorphic to $G/Z(G)$, quotient group to the kernel of the conjugation homomorphism. Maybe you meant this.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:08
$begingroup$
Also, it is considerably a bad idea to denote $Inn(G)$ by $N$, because just N suggests that it is a subgroup of G, but it isn't.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:24
$begingroup$
And also, to prove this result you don't have to do it directly by definition of normality (that it contains also all the conjugate elements). Because you must have strict distinction which elements are from $G$ and which are from $Aut(G)$. For example, your $g$, performing conjugation in $Aut(G)$ is also an automorphism (not necessarily inner), but not and element of $G$.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:26
$begingroup$
You write "And the whole image of a group homomorphism is always a normal subgroup in the codomain". This is false. Take, as a counterexample, any non-normal subgroup $H le G$ and consider the inclusion map $H to G$ given simply by $i(h) = h$.
$endgroup$
– Clay Thomas
Jun 27 '18 at 13:59
add a comment |
$begingroup$
First, you used additive notation for the group operation, but we needn't suppose that the group $G$ is abelian, the result holds in the general case.
Second, group automorphisms $Aut(G)$ is a different group from $G$ itself (so, inner automorphisms do not form a subgroup in $G$...).
The set/class of all inner automorphisms $Inn(G)$ form a normal subgroup in $Aut(G)$, not in $G$ itself!
Conjugation operation, you called it $f(x)$, more precisely $f_a(x)$ or $x^a$, is a group homomorphism $Grightarrow Aut(G)$ (not necessarily isomorphism or even mono- or epi-).
And the whole image of a group homomorphism is always a normal subgroup in the codomain (the proof is a bit verbose).
In this case the whole image of the conjugation operation is exactly the set of all inner automorphisms (forming a subgroup in $Aut(G)$).
$$ f: Grightarrow Aut(G) $$
$$ f_a(x) := a^{-1}xa $$
$$ f_G = Inn(G) $$
$$ Inn(G) triangleleft Aut(G) $$
So we can indeed always speak about outer automorphisms $Out(G)$ of a group $G$, which is possibly trivial, because $Inn(G)$ exhaust the whole $Aut(G)$ (all automorphisms are inner), or, conversely, equal (isomorphic) to $Aut(G)$, because $Inn(G)$ is trivial (in case of abelian groups).
answered Aug 22 '15 at 23:55
Andrew MiloradovskyAndrew Miloradovsky
37418
$endgroup$
First, you used additive notation for the group operation, but we needn't suppose that the group $G$ is abelian, the result holds in the general case.
Second, group automorphisms $Aut(G)$ is a different group from $G$ itself (so, inner automorphisms do not form a subgroup in $G$...).
The set/class of all inner automorphisms $Inn(G)$ form a normal subgroup in $Aut(G)$, not in $G$ itself!
Conjugation operation, you called it $f(x)$, more precisely $f_a(x)$ or $x^a$, is a group homomorphism $Grightarrow Aut(G)$ (not necessarily isomorphism or even mono- or epi-).
And the whole image of a group homomorphism is always a normal subgroup in the codomain (the proof is a bit verbose).
In this case the whole image of the conjugation operation is exactly the set of all inner automorphisms (forming a subgroup in $Aut(G)$).
$$ f: Grightarrow Aut(G) $$
$$ f_a(x) := a^{-1}xa $$
$$ f_G = Inn(G) $$
$$ Inn(G) triangleleft Aut(G) $$
So we can indeed always speak about outer automorphisms $Out(G)$ of a group $G$, which is possibly trivial, because $Inn(G)$ exhaust the whole $Aut(G)$ (all automorphisms are inner), or, conversely, equal (isomorphic) to $Aut(G)$, because $Inn(G)$ is trivial (in case of abelian groups).
answered Aug 22 '15 at 23:55
Andrew MiloradovskyAndrew Miloradovsky
37418
answered Aug 22 '15 at 23:55
Andrew MiloradovskyAndrew Miloradovsky
37418
answered Aug 22 '15 at 23:55
Andrew MiloradovskyAndrew Miloradovsky
37418
answered Aug 22 '15 at 23:55
Andrew MiloradovskyAndrew Miloradovsky
37418
37418
$begingroup$
PS: Actually, we should use pairing (carrier, oepration(s)) to denote groups without ambiguity, but we assume, as always, that the carrier and the algebraic structure are denoted by the same symbol.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:03
$begingroup$
Also, $Inn(G)$ is isomorphic to $G/Z(G)$, quotient group to the kernel of the conjugation homomorphism. Maybe you meant this.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:08
$begingroup$
Also, it is considerably a bad idea to denote $Inn(G)$ by $N$, because just N suggests that it is a subgroup of G, but it isn't.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:24
$begingroup$
And also, to prove this result you don't have to do it directly by definition of normality (that it contains also all the conjugate elements). Because you must have strict distinction which elements are from $G$ and which are from $Aut(G)$. For example, your $g$, performing conjugation in $Aut(G)$ is also an automorphism (not necessarily inner), but not and element of $G$.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:26
$begingroup$
You write "And the whole image of a group homomorphism is always a normal subgroup in the codomain". This is false. Take, as a counterexample, any non-normal subgroup $H le G$ and consider the inclusion map $H to G$ given simply by $i(h) = h$.
$endgroup$
– Clay Thomas
Jun 27 '18 at 13:59
add a comment |
$begingroup$
PS: Actually, we should use pairing (carrier, oepration(s)) to denote groups without ambiguity, but we assume, as always, that the carrier and the algebraic structure are denoted by the same symbol.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:03
$begingroup$
Also, $Inn(G)$ is isomorphic to $G/Z(G)$, quotient group to the kernel of the conjugation homomorphism. Maybe you meant this.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:08
$begingroup$
Also, it is considerably a bad idea to denote $Inn(G)$ by $N$, because just N suggests that it is a subgroup of G, but it isn't.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:24
$begingroup$
And also, to prove this result you don't have to do it directly by definition of normality (that it contains also all the conjugate elements). Because you must have strict distinction which elements are from $G$ and which are from $Aut(G)$. For example, your $g$, performing conjugation in $Aut(G)$ is also an automorphism (not necessarily inner), but not and element of $G$.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:26
$begingroup$
You write "And the whole image of a group homomorphism is always a normal subgroup in the codomain". This is false. Take, as a counterexample, any non-normal subgroup $H le G$ and consider the inclusion map $H to G$ given simply by $i(h) = h$.
$endgroup$
– Clay Thomas
Jun 27 '18 at 13:59
$begingroup$
PS: Actually, we should use pairing (carrier, oepration(s)) to denote groups without ambiguity, but we assume, as always, that the carrier and the algebraic structure are denoted by the same symbol.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:03
$begingroup$
PS: Actually, we should use pairing (carrier, oepration(s)) to denote groups without ambiguity, but we assume, as always, that the carrier and the algebraic structure are denoted by the same symbol.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:03
$begingroup$
Also, $Inn(G)$ is isomorphic to $G/Z(G)$, quotient group to the kernel of the conjugation homomorphism. Maybe you meant this.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:08
$begingroup$
Also, $Inn(G)$ is isomorphic to $G/Z(G)$, quotient group to the kernel of the conjugation homomorphism. Maybe you meant this.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:08
$begingroup$
Also, it is considerably a bad idea to denote $Inn(G)$ by $N$, because just N suggests that it is a subgroup of G, but it isn't.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:24
$begingroup$
Also, it is considerably a bad idea to denote $Inn(G)$ by $N$, because just N suggests that it is a subgroup of G, but it isn't.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:24
$begingroup$
And also, to prove this result you don't have to do it directly by definition of normality (that it contains also all the conjugate elements). Because you must have strict distinction which elements are from $G$ and which are from $Aut(G)$. For example, your $g$, performing conjugation in $Aut(G)$ is also an automorphism (not necessarily inner), but not and element of $G$.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:26
$begingroup$
And also, to prove this result you don't have to do it directly by definition of normality (that it contains also all the conjugate elements). Because you must have strict distinction which elements are from $G$ and which are from $Aut(G)$. For example, your $g$, performing conjugation in $Aut(G)$ is also an automorphism (not necessarily inner), but not and element of $G$.
$endgroup$
– Andrew Miloradovsky
Aug 23 '15 at 0:26
$begingroup$
You write "And the whole image of a group homomorphism is always a normal subgroup in the codomain". This is false. Take, as a counterexample, any non-normal subgroup $H le G$ and consider the inclusion map $H to G$ given simply by $i(h) = h$.
$endgroup$
– Clay Thomas
Jun 27 '18 at 13:59
$begingroup$
You write "And the whole image of a group homomorphism is always a normal subgroup in the codomain". This is false. Take, as a counterexample, any non-normal subgroup $H le G$ and consider the inclusion map $H to G$ given simply by $i(h) = h$.
$endgroup$
– Clay Thomas
Jun 27 '18 at 13:59
add a comment |
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3
$begingroup$
For clarity, instead of $f$, let $varphi_a$ denote the inner automorphism $varphi_a(x)=a^{-1}xa$. Then show $gvarphi_a g^{-1}=varphi_{g(a)}$, hence $gvarphi_a g^{-1}$ is inner.
$endgroup$
– Ben West
Aug 22 '15 at 22:04
4
$begingroup$
It's a normal subgroup of $operatorname{Aut}G$, not of $G$.
$endgroup$
– Bernard
Aug 22 '15 at 22:11
2
$begingroup$
Your title is deceptive. The question is about inner automorphisms from a group to itself, not to another group. In fact, one only uses the terminology "automorphisms" when talking about isomorphisms from an object to itself.
$endgroup$
– Lee Mosher
Aug 22 '15 at 22:33
$begingroup$
$g(a^{-1}xa)g^{-1} = ga^{-1}xag^{-1} = (ag^{-1})^{-1}x(ga^{-1})^1$ but I don't think this help. Can I argue that $ag^{-1}$ is in $G$?
$endgroup$
– Guerlando OCs
Aug 22 '15 at 23:43
1
$begingroup$
@GuerlandoOCs No, $g$ is an automorphism of $G$, not an element of $G$, so $$gvarphi_ag^{-1}(x)=gvarphi_a(g^{-1}(x))=g(a^{-1}g^{-1}(x)a)=g(a^{-1})g(g^{-1}(x))g(a)=g(a)^{-1}xg(a)=varphi_{g(a)}(x).$$
$endgroup$
– Ben West
Aug 23 '15 at 0:03