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If $G$ is Abelian group and $a,b \in G$ are distinct elements of order $2$, show that $ab$ has order $2$. Prove that $H=\{e,a,b,ab\}$ forms a subgroup of $G$ that is not cyclic.

I request help on how I can show that $ab$ has order $2$ as well as to show $H$ is not cyclic. I think there exist no $x \in G$ that generates $H$.

Thanks all.

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    $(ab)(ab) = aabb = ee = e$ to show $(ab)$ has order $2$. And to show that $H$ is not cyclic $a$ and $b$ are distinct and $a^2 = b^2 = e$ so $a^n \neq b$ for any $n$ and similarly $b^k \neq a$ for any $k$.2011-12-17
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    @DimitrijeKostic $a^{-1} = a$ since they have order 2 ... (so $b\neq a^{-1}$ as they are assumed distinct)2011-12-17
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    $G$ abelian implies $(ab)^2 = a^2 b^2$.2011-12-17

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