Induction on Groups with Cyclic Sylow 2-Subgroups (4.5.49)

Dummit and Foote Abstract Algebra, section 4.5, exercise 49:

Prove that if | G | = 2ⁿm where m is odd and P∈Syl₂(G) is cyclic then G has a normal subgroup of order m. [Use induction and exercises 11 and 12 in section 2.]

Work by induction, assuming that all groups with a cyclic Sylow 2-subgroup and orders of the form 2^zm for z < n have a normal subgroup of order m. The basis case for a group of order 2m has already been established in the aforementioned exercises.

Let P = < x >, so that x has an order of 2ⁿ. As made apparent in exercise 11, x's image under the regular left representation of G into S_G is an element comprised of m 2ⁿ-cycles. Since m is odd and the parity of an even-ordered cycle is odd, we have that x is an odd permutation. This implies that GA_G = S_G and by the Second Isomorphism Theorem, | S_G : A_G | = 2 = | G : G∩A_G |, so that G∩A_G is a group of order 2^n-1m. Note that x² is an element of order 2^n-1 within G∩A_G, so that Q = < x² > ∈ Syl₂(G∩A_G) is cyclic, and by the inductive assumption, there is a subgroup N of order m in G such that G∩A_G ≤ N_G(N), and the latter is either of order 2^n-1m or 2ⁿm, and in particular is either G∩A_G or G.

We now prove N char G∩A_G so that N is normal in G by a previous exercise. Assume that there is a distinct subgroup M of order m in G∩A_G so that | NM | = | N | | M | / | N∩M |, all of which are odd numbers, so | NM | is of odd order. Since N ≤ NM and N is the largest subgroup of odd order, we have N = NM and M = N, violating the premises. Therefore, N char G∩A_G, so N is normal in G.

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Clearly, the same technique used to prove N char G∩A_G can be applied, mutatis mutandis, to prove N char G.