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I am presented with the following problem:

Let $n \in \mathbb{N}$, let $G$ be a group of order $2n$, and let $H$ be a subgroup of order $n$. Prove that $gH = Hg$ for all $g \in G$.

My initial idea was to prove that $G$ is abelian, since if $G$ is abelian, then $H$ is abelian too ($H$ is a subgroup of $G$), which means that $gH = Hg$. However, I am having difficulty proving that $G$ is abelian just from the fact that its order is $2n$. If anybody could give me any hints in the right direction, that would be greatly appreciated.

Shaun
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Justin Trudeau
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  • Well it clearly does not work: $S_3$ is not abelian and has a subgroup of half order. – Arctic Char Oct 17 '21 at 11:48
  • $G$ might not be abelian. You can look at my answer here: https://math.stackexchange.com/questions/4266238/let-h-le-g-show-if-gh-g-h-2-then-h-unlhd-g-but-if-gh-3-the/4266244#4266244 – Mark Oct 17 '21 at 11:49
  • Please use more descriptive titles. – Shaun Oct 17 '21 at 12:01

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