Show that the order of any conjugacy class is a power of $p$.
Problem:
Let $N$ be a normal subgroup of $G$. Suppose $Aut(N)$ is a $p$-group for a prime $p$. Show that the order of any $G$-conjugacy class in $N$ is a power of $p$.
My attempt:
Let $G$ act on $N$ by conjugation, which induces a homomorphism $varphi:Gto Aut(N)$. By the first isomorphism theorem we then have that $G/ker(varphi) cong im(varphi) leq Aut(N)$. Thus I have that $[G:ker(varphi)]$ divides $p^k$. However, I'm having trouble relating this back to conjugacy classes. I know $ker(varphi)$ is the centralizer of $N$, that is, $C_G(N)$. I also know that $[G:C_G(g_i)]$ is the size of a conjugacy class with representative $g_i$. But I'm still missing something to put this all together.
Any help is appreciated.
abstract-algebra group-theory
asked Dec 10 '18 at 20:31
M47145
3,26331130
add a comment |
Problem:
Let $N$ be a normal subgroup of $G$. Suppose $Aut(N)$ is a $p$-group for a prime $p$. Show that the order of any $G$-conjugacy class in $N$ is a power of $p$.
My attempt:
Let $G$ act on $N$ by conjugation, which induces a homomorphism $varphi:Gto Aut(N)$. By the first isomorphism theorem we then have that $G/ker(varphi) cong im(varphi) leq Aut(N)$. Thus I have that $[G:ker(varphi)]$ divides $p^k$. However, I'm having trouble relating this back to conjugacy classes. I know $ker(varphi)$ is the centralizer of $N$, that is, $C_G(N)$. I also know that $[G:C_G(g_i)]$ is the size of a conjugacy class with representative $g_i$. But I'm still missing something to put this all together.
Any help is appreciated.
abstract-algebra group-theory
asked Dec 10 '18 at 20:31
M47145
3,26331130
1
The orbit of a point has size dividing the order of the group acting, namely $G/ker(phi)$. Put another way, $C_G(N)le C_G(n)$ for all $nin N$.
– Hempelicious
Dec 11 '18 at 0:45
add a comment |
Problem:
Let $N$ be a normal subgroup of $G$. Suppose $Aut(N)$ is a $p$-group for a prime $p$. Show that the order of any $G$-conjugacy class in $N$ is a power of $p$.
My attempt:
Let $G$ act on $N$ by conjugation, which induces a homomorphism $varphi:Gto Aut(N)$. By the first isomorphism theorem we then have that $G/ker(varphi) cong im(varphi) leq Aut(N)$. Thus I have that $[G:ker(varphi)]$ divides $p^k$. However, I'm having trouble relating this back to conjugacy classes. I know $ker(varphi)$ is the centralizer of $N$, that is, $C_G(N)$. I also know that $[G:C_G(g_i)]$ is the size of a conjugacy class with representative $g_i$. But I'm still missing something to put this all together.
Any help is appreciated.
abstract-algebra group-theory
asked Dec 10 '18 at 20:31
M47145
3,26331130
Problem:
Let $N$ be a normal subgroup of $G$. Suppose $Aut(N)$ is a $p$-group for a prime $p$. Show that the order of any $G$-conjugacy class in $N$ is a power of $p$.
My attempt:
Let $G$ act on $N$ by conjugation, which induces a homomorphism $varphi:Gto Aut(N)$. By the first isomorphism theorem we then have that $G/ker(varphi) cong im(varphi) leq Aut(N)$. Thus I have that $[G:ker(varphi)]$ divides $p^k$. However, I'm having trouble relating this back to conjugacy classes. I know $ker(varphi)$ is the centralizer of $N$, that is, $C_G(N)$. I also know that $[G:C_G(g_i)]$ is the size of a conjugacy class with representative $g_i$. But I'm still missing something to put this all together.
Any help is appreciated.
abstract-algebra group-theory
abstract-algebra group-theory
asked Dec 10 '18 at 20:31
M47145
3,26331130
asked Dec 10 '18 at 20:31
M47145
3,26331130
asked Dec 10 '18 at 20:31
M47145
3,26331130
asked Dec 10 '18 at 20:31
M47145
3,26331130
asked Dec 10 '18 at 20:31
M47145
3,26331130
3,26331130
1
The orbit of a point has size dividing the order of the group acting, namely $G/ker(phi)$. Put another way, $C_G(N)le C_G(n)$ for all $nin N$.
– Hempelicious
Dec 11 '18 at 0:45
add a comment |
1
The orbit of a point has size dividing the order of the group acting, namely $G/ker(phi)$. Put another way, $C_G(N)le C_G(n)$ for all $nin N$.
– Hempelicious
Dec 11 '18 at 0:45
1
1
The orbit of a point has size dividing the order of the group acting, namely $G/ker(phi)$. Put another way, $C_G(N)le C_G(n)$ for all $nin N$.
– Hempelicious
Dec 11 '18 at 0:45
The orbit of a point has size dividing the order of the group acting, namely $G/ker(phi)$. Put another way, $C_G(N)le C_G(n)$ for all $nin N$.
– Hempelicious
Dec 11 '18 at 0:45
add a comment |
1 Answer
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The inner automorphism group is a $p$-group, being a subgroup of a $p$-group. The conjugacy class has a transitive action by the inner automorphism group. By Burnside's theorem, the conjugacy class has order a power of $p$.
answered Dec 10 '18 at 20:38
Matt Samuel
37.2k63465
add a comment |
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The inner automorphism group is a $p$-group, being a subgroup of a $p$-group. The conjugacy class has a transitive action by the inner automorphism group. By Burnside's theorem, the conjugacy class has order a power of $p$.
answered Dec 10 '18 at 20:38
Matt Samuel
37.2k63465
add a comment |
The inner automorphism group is a $p$-group, being a subgroup of a $p$-group. The conjugacy class has a transitive action by the inner automorphism group. By Burnside's theorem, the conjugacy class has order a power of $p$.
answered Dec 10 '18 at 20:38
Matt Samuel
37.2k63465
add a comment |
The inner automorphism group is a $p$-group, being a subgroup of a $p$-group. The conjugacy class has a transitive action by the inner automorphism group. By Burnside's theorem, the conjugacy class has order a power of $p$.
answered Dec 10 '18 at 20:38
Matt Samuel
37.2k63465
The inner automorphism group is a $p$-group, being a subgroup of a $p$-group. The conjugacy class has a transitive action by the inner automorphism group. By Burnside's theorem, the conjugacy class has order a power of $p$.
answered Dec 10 '18 at 20:38
Matt Samuel
37.2k63465
answered Dec 10 '18 at 20:38
Matt Samuel
37.2k63465
answered Dec 10 '18 at 20:38
Matt Samuel
37.2k63465
answered Dec 10 '18 at 20:38
Matt Samuel
37.2k63465
37.2k63465
add a comment |
add a comment |
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The orbit of a point has size dividing the order of the group acting, namely $G/ker(phi)$. Put another way, $C_G(N)le C_G(n)$ for all $nin N$.
– Hempelicious
Dec 11 '18 at 0:45