Show that the order of any conjugacy class is a power of $p$.

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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.










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  • 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


















2














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.










share|cite|improve this question


















  • 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
















2












2








2


1





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.










share|cite|improve this question













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






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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
















  • 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












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 $p$.






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    1 Answer
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    active

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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 $p$.






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      1














      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 $p$.






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        1












        1








        1






        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 $p$.






        share|cite|improve this answer












        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 $p$.







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        answered Dec 10 '18 at 20:38









        Matt Samuel

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