Classification of groups of an order two times a prime

Statement

Suppose $p$ is a prime number, $| G | = 2 p$ . Then, $G ≅ Z_{2 p}$ , the cyclic group, or $G ≅ D_{2 p}$ , the dihedral group.

Related facts

Classification of groups of order a product of two distinct primes (A more general case)

Facts used

Proof

By Sylow's theorem the number of Sylow subgroups of order $p$ satisfies

$n_{p} \equiv 1 mod p$ , $n_{p} ∣ 2$ , so $n_{p} = 1$ .

Hence there is exactly one subgroup $H \leq G$ of order $p$ . It must be cyclic since any group of prime order is cyclic.

By A subgroup of index 2 in the group is normal, $H ⊴ G$ .

So $H = ⟨ h : h^{p} = e ⟩$ .

Also by Sylow's theorem, there is at least one subgroup $K \leq G$ of order $2$ . Take $k$ to be the element of $K$ that is not the identity.

Then $k h^{j} \notin H$ for $j \in {1, . . ., p - 1}$ since $k h^{j} = h^{l} ⟹ k = h^{l - j} ⟹ k$ is order $1$ or $p$ by Lagrange's theorem, which is a contradiction.

Similarly, $k h^{j} \notin K$ .

Hence the $2 p$ unique elements of $G$ are ${e, h, . . ., h^{p - 1}, k, k h, . . ., k h^{p - 1}}$ .

Since $H ⊴ G$ , $k h k \in H$

Thus $k h k h = h^{j}$ , for some $j \in {0, . . ., p - 1}$ .

Note $k h \neq e$ .

This can now be split into two cases:

Case 1: $j \neq 0$

If $j \neq 0$ , then $(k h)^{2} \neq e$ .

Thus $(k h)^{p} = k h (h^{j})^{p - 1} \neq e$ . Hence $k h$ must have order $2 p$ . Since if a group contains an element of its order it is cyclic, $G ≅ Z_{2 p}$ .

Case 2: $j = 0$

If $j = 0$ , we have $y x y = x^{- 1}$ , which means that

$G = ⟨ h, k : h^{p} = e, k^{2} = e, k h k = h^{- 1} ⟩$

Which is a definition of the dihedral group by its presentation.

Hence $G ≅ D_{2 p}$ .

Examples

The following are the five smallest orders which are classified by this result: 4, 6, 10, 14, 22.