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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?2012-06-13

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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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    @JasonDeVito: Indeed. $SO_2$ is abelian though. A non-abelian example with the mentioned property is the [quaternion group](http://en.wikipedia.org/wiki/Quaternion_group).2012-06-13