# $T(G)$ may not be a subgroup?

It is obvious that for an abelian group $G$; the set of all torsion elements: $$T(G)=\{x\in G|x^n=0 \text{, for some nonzero integer } n\}$$ is a subgroup of the group. I am asked to probe this fact when $G$ is not abelian. Thanks.

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Every element always has the same order as its inverse. So you should try to find a non-abelian group with elements $x$ and $y$ of finite order such that $xy$ has infinite order. – Mikko Korhonen Oct 6 '12 at 18:40
@m.k.: Yes, exactly. T(G) is not a subgroup when G is not abelian. I am thinking about your simplification of the problem. Thanks for doing that. – S. Snape Oct 6 '12 at 18:52
It's worth noting that we tend to denote a multiplicative identity by $1$, not $0$. – Cameron Buie Nov 16 '12 at 2:46

In $G=SL_2(\mathbb Z)$ observe that $$\left(\begin{matrix}0&1\\-1&0\end{matrix}\right)\cdot\left(\begin{matrix}0&-1\\1&1\end{matrix}\right)=\left(\begin{matrix}1&1\\0&1\end{matrix}\right)$$ where the factors on the left ae in $T(G)$ and their product is not.
Another example can be found in the infinite dihedral group which is generated by elements $x$ and $y$ such that $x^2 = 1$ and $x^{-1}yx = y^{-1}$. – Mikko Korhonen Oct 6 '12 at 19:06
Actually, my example was not that magic, knowing that $PSL_2(\mathbb Z)\cong C_2*C_3 = \langle x,y\mid x^2=y^3=1\rangle$. It is just nice to have the abstract concept of amalgamated product available in such a concrete matrix group. If you want m.k.'s example (infinite dihedral group) just as "concrete": It is the group of isometries of $\mathbb Z$, i.e. maps of the form $x\mapsto x+k$ or $x\mapsto k-x$. The latter (and identity) make up $T(G)$ but are no subgroup. – Hagen von Eitzen Oct 7 '12 at 9:52
As a matter of interest, the dihedral group of infinite order is a special case of an amalgam of finite groups, as is the group ${\rm PSL}(2,\mathbb{Z}).$ In fact the former group is the free product $\mathbb{Z}/2\mathbb{Z} * \mathbb{Z}/2\mathbb{Z}$ and ${\rm PSL}(2,\mathbb{Z})$ is the free product $\mathbb{Z}/2\mathbb{Z}* \mathbb{Z}/3\mathbb{Z}$. In general, if $A$ and $B$ are two finite groups such that $C = A\cap B$ is neither $A$ nor $B,$ then the amalgam $A*_{C} B$ is an infinite group which is generated by its elements of finite order. A good reference for the theory of amalgams is the book "Trees" by J-P. Serre.
To supplement Geoff's answer, we can give the presentation $$\langle a,b\mid a^m=b^n=(ab)^p=1\rangle,$$ and as long as $\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\le 1$, the group is infinite. – user641 Oct 7 '12 at 0:38