Using Lagrange's theorem, prove that a non-abelian group of order $10$ must have a subgroup of order $5$.


Let $G$ be a group of order $10$. By Lagrange's theorem, if there exist a subgroup $H$ of $G$, the $o(H)=1,2,5~ or~ 10$. Assume that there is no subgroup of order $5$.

Let, $e\neq a\in G$. Then we have $o(a)|o(G)$. If every element of $G$ is of order $2$, then $a^2=e$ i.e $a=a^{-1}$ for all $a\in G$. Then I can show that $G$ is abelian. This is a contradiction.

Therefore, every element of $G$ is not of order $2$. Thus, $o(a)=5 ~or~ 10$. If $o(a)=5$, then $H= \left< a \right>$ is a cyclic subgroup of order $5$.

I don't know the approach is correct or not.

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    $\begingroup$ Looks great, nice work! It doesn't seem you explicitly tackled the case $o(a) = 10$, but this isn't so bad - if $o(a)$ is $10$, then $G$ is cyclic, hence abelian, a contradiction. One other remark: you say "every element of $G$ is not of order $2$". In fact, I think this should say "not every element of $G$ is of order $2$''. $G$ can have elements of order $2$ (indeed, take $D_{5}$, the symmetry group of the pentagon). However, it cannot be the case that ALL nonidentity elements of $G$ are of order $2$. $\endgroup$ – Alex Wertheim Oct 2 '15 at 5:20
  • $\begingroup$ @AlexWertheim Is there any other way to solve the problem. $\endgroup$ – user1942348 Oct 2 '15 at 5:32
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    $\begingroup$ Well, I'll precede this by saying that you've likely solved things in the intended way. But you could also use Cauchy's theorem to deduce the existence of an element of order $5$ directly. Cauchy's theorem is a kind of partial converse to Lagrange's theorem. If you don't know it/haven't covered it yet, you might consider looking it up. I wouldn't use it for this problem though; the way you've done things is optimal with the instructions given. $\endgroup$ – Alex Wertheim Oct 2 '15 at 5:35
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    $\begingroup$ A small note: after you find the contradiction (contradiction of the assumption that for every $a $ in $G $, o$(a)=2$), then you would say, $ \mathbf{there \, \, exists} $ at least one $a$ in $G$ such that o$(a)\neq 2 $. $\endgroup$ – Nizar Oct 2 '15 at 8:25

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