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.
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.
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?