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Let $H$ and $K$ be two subgroups of $G$ of finite order. Can $HK$ be infinite?

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Yes. – jspecter Sep 28 '12 at 15:46
I think there is a problem of interpretation here. From your title, it seems you are looking at the subgroup of $G$ generated by $HK$, which, in general, is not equal to $HK$. – M Turgeon Sep 28 '12 at 15:47
up vote 5 down vote accepted

This will depend on what you mean by $HK$.

If you mean just those elements in $G$ of the form $hk$ for $h \in H$, $k \in K$, then $|HK| \le |H||K|$.

If you mean the group generated by products of elements in $H$ and $K$, then the infinite dihedral group $\langle x,y | x^2=y^2=1 \rangle$ is an example. Take $H$ to be cyclic generated by $x$ and $K$ by $y$, then $G=HK$ and $|H|=|K|=2$.

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Let $G$ be the group of words made up of the characters 0 and 1, such that no character appears twice in a row. We define the group operation to be concatenation and also $0^2 = 1^2 = \epsilon$ where $\epsilon$ is the empty word.

Let $H$ be the subgroup $\{ \epsilon, 0 \}$ and $K$ be the subgroup $\{ \epsilon, 1 \}$. Then $G = HK$.

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Yes. Consider in the euclidean plane the reflection along the $x$ axis and the reflection along some other axis at an angle $\alpha$ with $\frac\alpha\pi\notin\mathbb Q$. These reflection generate a group that also contains infinitely many rotations by multiples of $2\alpha$.

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Remark: My group is isomorphic to Yoni Rozenshein's group. – Hagen von Eitzen Sep 28 '12 at 15:48

Ok, since it seems that the general consensus is to look at the subgroup generated by $HK$, here is another example.

The modular group $SL_2(\mathbb Z)$ is generated by two matrices of finite order: $$\begin{pmatrix} 0&1\\-1&0\end{pmatrix},\text{ and }\begin{pmatrix} 0&1\\-1&0\end{pmatrix}\begin{pmatrix} 1&1\\0&1\end{pmatrix}=\begin{pmatrix} 0&1\\-1&-1\end{pmatrix};$$ the former has order $4$ and the latter, order $3$. And of course, $SL_2(\mathbb Z)$ is infinite.

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