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Let $G$ be the orthogonal subgroup $O_2$. Show that the set $\{g \in G : g^2= e\}$ is not a subgroup of $G$

The question before says let $G$ be an abelian group and I can see where I have used that fact. It lets us write $a^2b^2=(ab)^2$ and so $ab \in G$. But I cant find a way to 'get out' of $G$.

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HINT: If $g$ and $h$ are elements of the set in question, is it true that the product $gh$ is in there? –  Joe Johnson 126 Jun 13 '12 at 13:45
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2 Answers

Hint: Every rotation is a product of two reflections.

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Somewhere in your proof you have used that $(gh)^2=g^2h^2.$ You need $G$ to be abelian for this.

In general $(gh)^2=ghgh$ and you cannot swap the inner $h$ and $g$ unless $G$ is abelian.

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Thanks think its time to take a break. This shouldn't have been that hard ;-). –  Ben Davidson Jun 13 '12 at 13:57
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Of course, $(gh)^2 = ghgh \neq g^2h^2$ is generally true, but you still need to find an explicit example of elements in $G$ with $(gh)^2\neq g^2h^2$. Other wise, it might have just happened on accident. For example, if instead of $O_2$, one used $SO_2$, then the corresponding $G$ would be a group. –  Jason DeVito Jun 13 '12 at 14:29
    
@JasonDeVito: Indeed. $SO_2$ is abelian though. A non-abelian example with the mentioned property is the quaternion group. –  Dejan Govc Jun 13 '12 at 14:59
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