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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.

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  • 2
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Apr 13 '13 at 7:25
  • $\begingroup$ For group theory without using Sylow theorems see the first line of Keith Conrad's notes. $\endgroup$ – Dietrich Burde Aug 18 '16 at 13:41
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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.

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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 $o(x)=p$.

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By cauchy's theorem for every prime dividing order of the group there exists an element of order p. proof of cauchy's theorem requires no sylow theorems. consider the subgroup generated by that element to get a subgroup of order n.

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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.

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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?

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