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I'm having some trouble with this question from a practice exam.

Let $G$ be finite group. Suppose for every $g \in G$ other than the identity element $e$, there is a subgroup $H \subset G$ of index $2$ that does not contain $g$.

Show that $g^2 = e$ for all $g \in G$.

My attempt:

Suppose $g \in G$. If $g = e$, obviously $g^2 = e$, so we can assume $g \not= e$. Let $H$ be the subgroup such that $|G:H| = 2$ and $g \not\in H$, which exists by assumption. Since left cosets and right cosets partition $G$, we must have $gH = Hg$, so $H$ is normal.

Since $g^2 \not\in gH$. $g^2H$ and $gH$ are two distinct cosets and partition $G$, and $g^2H = H$. Thus, $H = gHg$. Does this imply that $g = g^{-1}$? I don't think it does, but it looks a bit like conjugation, so maybe there's a theorem that I don't know about.

Or should I be following a different approach?

ppham27
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    This is false. For example in $S_n$, $A_n$ has index 2. So if this were true, every odd permutation would have order $2$. (It is however true that $g^2$ is in the identified subgroup $H$ of index 2, and that $g^2H$ is the identity in $G/H$.) – 2'5 9'2 Aug 20 '14 at 01:06
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    @alex.jordan What is the subgroup of index $2$ corresponding to an even permutation in your example? – angryavian Aug 20 '14 at 01:09
  • @angryavian I'm saying that if $g$ is odd, say a $4$-cycle, then we can find a subgroup of index $2$ that does not contain $g$: $A_n$. And yet $g^2$ is not the identity. – 2'5 9'2 Aug 20 '14 at 01:09
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    @alex.jordan But I thought the hypothesis is that for each element of the group there exists a corresponding subgroup, not just for some elements. – angryavian Aug 20 '14 at 01:10
  • Ah, that's what I get for reading the question from the title and not the body. – 2'5 9'2 Aug 20 '14 at 01:10

1 Answers1

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Does this work? I may have solved it myself.

Since such a subgroup exists for every $g \in G$ such that $g \not= e$. Suppose for a contradiction $g^2 \not= e$. Let $K$ be the subgroup such that $[G:K] = 2$ and $g^2 \not\in K$. $g \not\in K$, either because that would imply that $g^2 \in K$. Thus, $K$ and $gK$ partition $G$. $K$ and $g^2K$ also partition $G$. Then, $gK = g^2K \Leftrightarrow g \in K$, which is a contradiction. Therefore, $g^2 = e$ must be true.

ppham27
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