A finite group with a cyclic Sylow 2-subgroup has a normal subgroup of maximal odd order

Prove that if |G| = 2^nm where m is odd and G has a cyclic Sylow 2-subgroup then G has a normal subgroup of order m.


We proceed by induction on n.

For the base case, suppose |G| = 2m where m is odd. Note that every Sylow 2-subgroup is cyclic. By this previous exercise, G has a subgroup of index 2, hence order m, which is necessarily normal.

For the inductive step, suppose that for some n \geq 1 and all odd m, any group of order 2^nm which has a cyclic Sylow 2-subgroup has a normal subgroup of order m. Let G be a group of order 2^{n+1}m and with a cyclic Sylow 2-subgroup. Let x \in G be a generator of the Sylow 2-subgroup; i.e. |x| = 2^{n+1}. Let \varphi : G \rightarrow S_G be the left regular representation of G; by this previous exercise, \varphi(x) is an odd permutation. By this previous exercise, G has a subgroup H of index 2, which is necessarily normal in G. Note that all Sylow 2-subgroups of G are cyclic, and that every Sylow 2-subgroup of H is contained in a Sylow 2-subgroup of G. Thus H has a cyclic Sylow 2-subgroup, in particular H \cap \langle x \rangle = \langle x^2 \rangle since \langle x \rangle has a unique subgroup of order 2^n. By the induction hypothesis, H has a normal subgroup N of order m. Note that H \leq N_G(N) \leq G, and that since H is maximal, N_G(N) \in \{ H,G \}.

We claim that N is the unique subgroup of H of order m. To prove this claim, suppose to the contrary that M \leq H is a subgroup of order m. Now NM \leq H is a subgroup and strictly contains N and M. Moreover, we have |NM| = |N| \cdot |M| /|M \cap N|, so that by Lagrange, |NM| is odd. This is a contradiction, though, because m is the largest odd integer dividing |H|. Thus N is the unique subgroup of order m. Hence N is characteristic in H. By this previous exercise, then, N is normal in G.

By induction, the conclusion holds for all groups of order 2^nm having a cyclic Sylow 2-subgroup.

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