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I am being involved with the following problem:

A group $G$ of order $2^{k}m$ wherein $m$ is odd has a cyclic subgroup of order $2^k$. Prove that $G$ has a normal subgroup of index $2^k$.

Honestly, I have one solution of this problem which is based on induction on $k$ and permutation groups. I am just asking if someone knows any other approaches. Thans

Shaun
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Mikasa
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  • @Cocopuffs: A Sylow subgroup would be of order $2^k$, not of index $2^k$. – Najib Idrissi Jun 25 '12 at 13:46
  • @N.I Whoops, I didn't catch that – Cocopuffs Jun 25 '12 at 13:46
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    Another (but less elementary) approach is to use Burnside's Transfer Theorem. – Derek Holt Jun 25 '12 at 16:28
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    @DerekHolt: I could find the Theorem as follows. Sign me if it is the right one. if $G$ is a finite group, $P$ a Sylow subgroup, and $N(P) = C(P)$, then $G$ contains a normal subgroup $N$ with $NP = G$ and $N$ intersect $P$ $= 1$, i.e. $G$ is a semidirect product of $P$ with $N$. Thanks for visiting my question – Mikasa Jun 25 '12 at 17:20
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    @Babak, yes your $N$ is called a normal p-complement in this case. Observe that the requirement $N_G(P)=C_G(P)$ is equivalent to $P \subseteq Z(N_G(P))$. $P$ must be abelian to satisfy the condition. Now in general, if $p$ is the smallest prime dividing $|G|$, and a Sylow $p$-subgroup of $G$ is cyclic, then the condition is satisfied (this follows from the fact that $N_G(P)/C_G(P)$ injects in $Aut(P)$!) and hence $G$ has a normal $p$-complement. There are many $p$-complement theorems of which the one of Thompson is most famous: it implied that kernels of Frobenius groups are nilpotent. – Nicky Hekster Jun 25 '12 at 20:49
  • @NickyHekster: Thanks Nicky. I really need your detailed comment here. It sparked by Prof. Holt and thanks both of you. – Mikasa Jun 26 '12 at 07:55

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Consider the regular representation $\rho_{G}$ of $G$. Let $S$ be a Sylow $2$-subgroup of $G,$ and suppose that $S = \langle s \rangle$ is cyclic. Then $\rho_{G}$ restricts to $S$ as a drect sum of $m$ copies of $\rho_{S}.$ Hence the eigenvalues of $\rho_{G}(s)$ all occur with multiplicity $m,$ and each $2^{k}$-th root of unity occurs as such an eigenvalue. Now the product of all complex $2^{k}$th roots of unity is $-1$, because all such roots except $1$ and $-1$ occur in complex conjugate pairs. Hence $\rho_{G}(s)$ has determinant $(-1)^{m} = -1$ as $m$ is odd. Now $g \to {\rm det}(\rho_{G}(g))$ is a $1$-deimensional representation of $G.$ If $g$ has odd order, then ${\rm det}(\rho_{G}(g)) =1,$ because all $|g|$-th rooots of unity occur as eigenvalues with equal multiplicity, and all except 1 occur in complex conjugate pairs. Hence the image of this $1$-dimensional representation is $\{1,-1\}$ (I have skipped a few details here), and its kernel has index $2$. By induction, the kernel has a normal subgroup of index $2^{k-1}$ which is characteristic, so a normal subgroup of index $2^{k}$ of $G.$

Geoff Robinson
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  • Sorry, I don't see why the subgroup of $G$ is characteristic. You used it in your induction but didn't show the subgroup of index $2^k$ is characteristic. Could you please specify? – Juggler Nov 22 '20 at 03:28
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    The point is that if there is a normal subgroup $H$ of order $m$, then $H$ contains all elements of $G$ of odd order ( but no element of even order by Lagrange's Theorem). Hence $H$ (as a set) is just the set of all elements of $G$ of odd order, and as such, is invariant under any automorphism of $G$, so is characteristic. – Geoff Robinson Nov 22 '20 at 09:59