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. $\Box$