# Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)

I want to show without using Sylow theorem that

Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)

My attempt: Since $|G|=2p,$ even $\exists~a\neq e\in G$ such that $a^{-1}=a.$ Then $H=\{e,a\}\leq G.$ Let $S$ be the set of all left cosets of $H$ in $G.$ Now $S$ forms a group w.r.t. the composition $g_1Hg_2H=(g_1g_2)H.$ The mapping $\sigma:S\to G:gH\mapsto g$ is an $1-1$ homomorphism. Then $S\simeq\sigma(S)\leq G.$ Thus $G$ has a subgroup of order $p.$

Please tell me whether I'm right!

I'm skecptical as I didn't use that $p$ is prime.

• Two problems in your attempt. 1) The multiplication of cosets of $H$ is not always well defined. 2) The mapping $\sigma$ is not well defined. – Jyrki Lahtonen Apr 13 '13 at 7:25
• For group theory without using Sylow theorems see the first line of Keith Conrad's notes. – Dietrich Burde Aug 18 '16 at 13:41

We only need to consider the case that $p$ is odd.If there is no element of order $p$ in the group, then every non-identity element of the group has order $2$. But a group in which every non-identity element as order $2$ is Abelian: ($1 = abab,$ so $aba = b$ and $ba = ab).$ The group has more than one subgroup of order $2,$ so since it is Abelian, it has a subgroup of order $4,$ contrary to Lagrange's theorem.
Write $G$ for the group. If $G$ is cyclic, let $g$ be a generator, and we have that $g^2$ has order $p$. Otherwise, note that $G$ is solvable by Burnside's theorem. The only simple solvable groups are cyclic of order $p$, so we may write $1\lhd N \lhd G$. If $N$ has order $p$, we're done. If $N$ has order $2$, then take a nontrivial element $xN$ in $G/N$. Then $o(xN)$ divides $o(x)$, but $G$ is not cyclic, so it follows that $o(x)=p$.
First why that subgroup of order two is normal? every group of order 2p need not have a normal subgroup of order 2. Consider $S_3$. So in your proof S need not form a group in general under coset product.
Why the set of left cosets of $H$ forms a group? The question is a special case of Every group has a subgroup of prime order?