Problem 2. Let G be a group with |G|=2^nm, where m is odd. Suppose that G has a cyclic Sylow 2-subgroup. Prove that G has a normal subgroup of order m.

Solution. Let P be a cyclic Sylow 2-subgroup. We solve the problem using induction over n, the exponent of 2 in |G|. There is nothing to prove if n=0. If n=1, then we are done by Problem 1. Now suppose n \geq 2. By Problem 1, G has a normal subgroup N of index 2, i.e. |N|=2^{n-1}m. By Lemma 2, N \cap P is a Sylow 2-subgroup of N. Clearly N \cap P is cyclic because it is a subgroup of the cyclic group P. Thus, by the induction hypothesis, N has a subgroup K such that K is normal in N and |K|=m. So we only need to show that K is normal in G. We will do this in two steps:

Step 1. K is the unique subgroup of order m in N. To see this, suppose L is another subgroup of order m in N. Then KL is a subgroup of N, because K is normal in N. Thus |N|=2^{n-1}m=r|KL|, for some integer r. But |KL|=\frac{m^2}{|K \cap L|} and so 2^{n-1}|K \cap L|=mr, which gives us m \mid |K \cap L|, because m is odd. But |K \cap L| \leq |K|=m and so |K \cap L|=m=|K|=|L|. Therefore K = L.

Step 2. K is normal in G. To see this, let g \in G and put L=gKg^{-1}. Then, since N is normal in G, we have  L \subseteq gNg^{-1}=N. So K and L are both subgroups of N and |L|=|K|. Hence K=L, by step 1. \Box

Corollary 1. Non-abelian groups of order 4k + 2 are not simple.

Proof. Let G be a non-abelian group of order 4k+2. Then a Sylow 2-subgroup of G has order 2 and so it is cyclic. So, by Problem 2, G has a normal subgroup of order 2k+1. \ \Box

We can extend the result in Problem 2 to a much stronger one:

Problem 3. Let G be a group with |G|=p^nm, where p is the smallest prime divisor of |G| and \gcd(m,p)=1. Suppose that G has a cyclic Sylow p-subgroup. Prove that G has a normal subgroup of order m.

Solution. Let P be a cyclic Sylow p-subgroup. Let C(P) and N(P) be the centralizer and the normalizer of P in G, respectively. By the lemma in this post, N(P)/C(P) is isomorphic to a subgroup of \text{Aut}(P). Since |P|=p^n and P is cyclic, we have

|\text{Aut}(P)|=p^{n-1}(p-1).

Thus |N(P)/C(P)| \mid p^{n-1}(p-1). But since |N(P)/C(P)| \mid |G| and p is the smallest prime divisor of |G|, we have

\gcd(|N(P)/C(P)|, p-1)=1.

Thus |N(P)/C(P)| must divide p^{n-1}.  So if |N(P)/C(P)| > 1, then p \mid |N(P)/C(P)| and hence, since P \subseteq C(P), we will have p^{n+1} \mid |N(P)| which is absurd. Therefore N(P)=C(P) and the result now follows from the Burnside’s Normal Complement Theorem. \Box

Corollary 2. If G is a non-abelian group and |G|=pm, where p is the smallest prime divisor of |G| and \gcd(p,m)=1, then G is not simple.

Proof. Every Sylow p-subgroup of G has order p and so it is cyclic. Thus, by Problem 3, G has a normal subgroup of order m and so G is not simple. \Box

Corollary 3. Non-abelian groups of square-free order are not simple. \Box

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