Classification of groups of an order two times a prime

Statement

Suppose $p$ is a prime number, $|G|=2p$ . Then, $G\cong \mathbb {Z} _{2p}$ , the cyclic group, or $G\cong D_{2p}$ , 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

$n_{p}\equiv 1{\bmod {p}}$ , $n_{p}\mid 2$ , 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, $H\trianglelefteq G$ .

So $H=\langle h:h^{p}=e\rangle$ .

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

Similarly, $kh^{j}\notin K$ .

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

Since $H\trianglelefteq G$ , $khk\in H$

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

Note $kh\neq e$ .

This can now be split into two cases:

Case 1: $j\neq 0$

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

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