# Product of subgroups which is not a group - infinite case

It is well known that a product of subgroups might not be a subgroup. I give an example here.

Would you have an example of an infinite group having two infinite subgroups whose product is not a group? Better to have a case not homeomorphic to a finite case.

Also about an example of two subgroups which are not normal and whose product is a proper subgroup?

• what do you mean by not homeomorphic? – user225222 Apr 12 '15 at 11:28
• @user225222 it means that you can't find a group $G^\prime$ and a group homeomorphism $\varphi$ from $G$ to $G^\prime$ such that $\varphi(G)$ is finite. In the case you provide below $\varphi: G \times G_1$ with $\varphi(g,g1) \mapsto (g,1)$ is such homeomorphism. – mathcounterexamples.net Apr 12 '15 at 12:41

1. Straighforward: Let $G$ be free with two generators $a,b$ and let $H=\langle a\rangle$, $K=\langle b\rangle$.

2. Let $G=S_5$, $H=A_4$, $K=\langle (1\,2\,3\,4\,5)\rangle$. Then $HK=A_5$

A perhaps easier example for a product of two non-normal subgroups being a proper subgroup:

$$G=A_4\;,\;\;K=\{(1)\,,\,(12)(34)\}\;,\;\;H=\{(1)\,,\,(13)(24)\}$$

Thus,

$$\;H,K< G\;,\;\;H,K\rlap{\;\;/}\lhd G\;,\;\;HK=KH<G\;$$

BTW, it happens to be that $\;HK\lhd G\;$ ...:)

1)Let G, H,K be the groups of your example such that $H,K\subset G$ and $HK$ is not a group. Let $G_1$ be an infinite group. $H_1:=H\times G_1,K_1:=K\times G_1$ are subgroups of $G\times G_1$. And there product is not a subgroup of $G\times G_1$.

2) Let $H$ be a non normal abelian subgroup of $G$. $HH$ is a proper group. Indeed, since $H$ is abelian $HH$ is a group (group of squares) and it's proper because it's a subset of $H$ who is proper. Permutation example: $G=\mathfrak{S}_3$, $H=\{1,(1,2)\}$

• What is an "ur" example? – bof Apr 12 '15 at 11:46
• @bof "The question says : It is well known that a product of subgroups might not be a subgroup. I give an example here." So ur example is the example given by jean-pierre. I haven't read it – user225222 Apr 12 '15 at 11:50
• @bof why are u asking? – user225222 Apr 12 '15 at 11:55
• @bof: "ur" is a (quite common) misspelling of "your" – celtschk Apr 12 '15 at 12:07

Take the group of rotations in 3D space, $\mathrm{SO}(3)$, and take the subgroups of rotations around the $z$ axis (let's call that $S_z$), and of rotations around the $x$ axis ($S_x$). Both subgroups are clearly infinite, but the product doesn't commute (for example, $S_zS_x$ contains no element that maps $e_x$ to $e_z$, but $S_xS_z$ does ($S_z$ allows to rotate $e_x$ to $e_y$, and then $S_x$ allows to rotate $e_y$ to $e_z$), thus as you proved on the linked site, the product is not a subgroup.

Let $G=\text{GL}(2,\mathbb{R})$, let $$H=\left\langle \begin{pmatrix}1&0\\0&2\end{pmatrix}\right\rangle;\ K=\left\langle\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}1&0\\0&2\end{pmatrix}\begin{pmatrix}1&1\\1&-1\end{pmatrix}^{-1}\right\rangle$$
You can compute that $$HK=\left\{\frac{1}{2}\begin{pmatrix}1+2^m&1-2^m\\2^n-2^{m+n}&2^n+2^{m+n}\end{pmatrix}\ s.t.\ m,n\in\mathbb{Z}\right\}$$
$$KH=(HK)^{t}\neq HK$$ since $HK\setminus\text{Sym}(2,\mathbb{R})\neq\emptyset$