While studying for the algebra qual, I stumbled upon the following problem:

Let $G$ be a group of order $2p$ where $p$ is an odd prime. Prove that $G$ contains a nontrivial normal subgroup.

The problem is simple enough, however, it uses many results from group theory. We’ll investigate the problem in detail bellow.

Cosests and Lagrange’s Theorem

Definition. Let $H$ be a subgroup of a group $G$. The subset $aH=\{ah:h\in H\}$ of $G$ is the left coset of $H$ containing $a$. The subset $Ha=\{ha:h\in H\}$ is the right coset of $H$ containing $a$.

Cosets are important because they allow us to define normal subgroups, which in turn allow us to construct factor groups. We will need two important results concerning cosets.

If $H\leq G$, then the set of left cosets of $H$ form a partition of $G$. Indeed, let the relation $\sim_L$ be defined on $G$ by $a\sim_L b$ if and only if $a^{-1}b\in H$. It is easily verified that $\sim_L$ is an equivalence relation on $G$ with equivalence class $aH$. Moreover, $aH=H$ if and only if $a\in H$. This result follows a simple set containment argument along with the definition of left coset. It should be noted that both of these results have an analogue for right cosets.

Lagrange’s Theorem. If $G$ is a finite group and $H$ is a subgroup of $G$, then the order of $H$ divides the order of $G$.

Proof. Let $|G|=m$ and $|H|=n$. I claim that every left coset of $H$ has order $n$. Define the function $\varphi:H\to gH$ by $\varphi(h)=gh$. Then $\varphi$ is surjective by the definition of $gH$ as $\{gH:h\in H\}$. Suppose that $\varphi(h_1)=\varphi(h_2)$ for $h_1$ and $h_2$ in $H$. Then $gh_1=gh_2$ and by the cancellation law of a group, $h_1=h_2$. Thus, $\varphi$ is injective. Since $\varphi$ is a bijection, we conclude that $H$ and $gH$ have the same order and the claim is established. Since the left cosets of $H$ form a partition of $G$, we have that

$G = g_1H \cup g_2H \cup \cdots \cup g_rH$.

Thus,

$|G| = |g_1H| + |g_2H| + \ldots + |g_rH|$.

Since the order of each left coset of $H$ is equal to $n$, we have that

$|G| = m = n + n + \ldots + n$

where $n$ is summed up $r$ times. That is, $m=rn$ which shows that $n$ is a divisor of $m$. $\Box$

In our proof, to establish that every left coseet of $H$ has the same order as $H$ we showed that the function $\varphi(h)=gh$ was a bijection. It turns out that every right coset of $H$ also has the same order as $H$. To establish this, check that $\varphi:H\to Hg$ defined by $\varphi(h)=hg$ is a bijection.

Lagrange’s theorem is perhaps one of the most useful results in group theory since it allows us to actually compute orders of subgroups. The converse of Lagrange’s theorem is not true. The classic counter example is the alternating group $A_4$, which is the group of even permutations on $4$ elements. The order of $A_4$ is $12$ and $6$ is a divisor of $12$, but $A_4$ has no subgroup of order $6$. There are however some partial converses to Lagrange’s theorem. A partial converse which holds for arbitrary finite groups is the following result.

Cauchy’s Theorem. If $G$ is a finite group and $p$ is a prime dividing $|G|$, then $G$ has an element of order $p$.

Cauchy’s theorem will be useful for us in proving our original problem.

Normal Subgroups

Definition. A subgroup $N$ of a group $G$ is normal if $gN=Ng$ for all $g\in G$. If $N$ is a normal subgroup of $G$ then we shall write $N\unlhd G$.

The next theorem is of interest to us since it will be essential for proving our original problem.

Theorem 1. Let $G$ be a group of order $2n$. If $H$ is a subgroup of $G$ and $|H|=n$, then $H\unlhd G$.

Proof. Let $a\in G\setminus H$. Then $aH\neq H$, $Ha\neq H$, and $|aH|=|Ha|=n$. Since $|G|=2n$, we have that $G=H\cup aH$ and $G=H\cup Ha$. The elements of cosets are distinct because they from a partition of $G$, thus, it must be the case that $aH=Ha$. $\Box$

The Original Problem

Let $G$ be a group of order $2p$ where $p$ is an odd prime. Prove that $G$ contains a nontrivial normal subgroup.

Proof. By Cauchy’s theorem, there exists an element $a\in G$ such that $|a|=p$. Then the cyclic subgroup $\langle a \rangle$ is nontrivial and has order $p$. By theorem 1, $\langle a \rangle$ is a normal subgroup of $G$. $\Box$

The proof is quite elegant, but as you have seen, we needed to recall some important results in order achieve such a simple proof.