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.


|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.