# A nice group theory result

While working on some group theory problems today a friend and I came up with the following result.

Lemma. Let $H$ be a normal subgroup of a finite group $G$ such that $\gcd(|H|,|G/H|)=1$. If the order of $g\in G$ divides $|H|$, then $g\in H$.

Proof. Let $d$ be the order of $g$. Then the order $d'$ of $gH$ in $G/H$ divides both $d$ and $|G/H|$. But $\gcd(d,|G/H|)=1$. Hence $d'=1$, i.e., $gH=H$, i.e., $g\in H$. $\square$

Corollary 1. Let $H$ be a normal subgroup of a finite group $G$ such that $\gcd(|H|,|G/H|)=1$. If $K\le G$ such that $|K|$ divides $|H|$, then $K\le H$.

Proof. Apply the lemma to the elements of $K$. $\square$

Corollary 2. Let $H$ be a normal subgroup of a finite group $G$ such that $\gcd(|H|,|G/H|)=1$. Then $H$ is the unique subgroup of $G$ of order $|H|$.

Proof. Use Corollary 1. $\square$

Here is an example of the lemma in action.

Problem. Show that $S_4$ has no normal subgroup of order $8$ or $3$.

Solution. If $H$ is a normal subgroup of $S_4$ of order $8$, then $\gcd(|H|,|S_4/H|)=1$. Hence every element of order $2$ or $4$ in $S_4$ must lie in $H$. In particular, $(1\ 2),(1\ 2\ 3\ 4)\in H$. By a result in the previous post, $H=S_4$, a contradiction.

Likewise, if $H$ is a normal subgroup of $S_4$ of order $3$, then $H$ must contain every 3-cycle; in particular, $(1\ 2\ 3),(2\ 3\ 4)\in H$. Hence $(1\ 2\ 3)(2\ 3\ 4)=(1\ 2)(3\ 4)\in H$. But this has order 2, and $2\nmid 3$, a contradiction. $\square$

More generally, we can prove the following.

Corollary 3. $S_n$ for $n\ge 4$ has no non-trivial proper normal subgroup $H$ with $\gcd(|H|,|S_n/H|)=1$.

Proof. Suppose otherwise and let $d$ divide $|H|$. Then $H$ must contain all $d$-cycles. So if $|H|$ is even then taking $d=2$ gives $H=S_n$. If $|H|$ is odd, it contains the cycles $\sigma=(1\ \cdots\ d)$ and $\rho=(d\ \cdots\ 2\ n)$ for some $3\le d. Then $\sigma\rho=(1\ 2\ n)\in H$ has order 3. So $|H|$ contains all 3-cycles, i.e., $A_n\le H$. Since $A_n\le S_n$ is maximal, either $H=A_n$ or $H=S_n$, a contradiction. $\square$