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Let $G$ be a finite group of order $2n$ such that half of the elements of $G$ are of order $2$ and the other half form a subgroup $H$ of order $n$.

Then I know that $H$ is of odd order because for each $x \ne e$ in H, we have $x \ne x^{-1}$; thus after pairing all such elements we are left with the identity.

Also, the subgroup $H$, being of index two, is a normal subgroup of $G$.

How to determine if $H$ is abelian or not?

Saaqib Mahmood
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3 Answers3

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Hint: If $a^2=1$ then $aha^{-1}\cdot h=(ah)^2$.

Hagen von Eitzen
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  • Hagen von Eitzen, please expand upon your own hint. I'm sorry but I still haven't been able to figure out how to make anything of it. – Saaqib Mahmood Jul 06 '14 at 13:44
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Try to prove that G is isomorphic to $D_n$ (try to prove if you have an element of order 2,a, and one of order n,b, that generate all the group and $(ab)^2=1$ ) If you prove that, then H is generated by only one element of order n and it's abelian and cyclic

Patricio
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Try to show that $x \mapsto x^{-1}$ is an automorphism of $H$.

Shripad Garge
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  • Shripad Garge, how to show this? I know that once we have shown this, then we can easily show that $H$ is abelian. – Saaqib Mahmood Jul 06 '14 at 07:31
  • To prove this mapping is an automorphism, you have to show it is an operation preserving bijection from $H$ to $H$. – Oiler Oct 24 '15 at 22:57
  • Oh, I see that this question is still unanswered. Let me write more details. I am not sure how to create a math environment here so I will try to explain in words as much as possible.

    For any $a$ not in $H$, we have that $h \mapsto aha^{-1}$ is an automorphism of $H$. Now, since $a$ is of order two and so is the element $ah$, we get that $aha^{-1} = h^{-1}$. This proves that $h \mapsto h^{-1}$ is an automorphism of the group $H$.

     – Shripad Garge Aug 14 '17 at 10:22