Torsion Subgroup (Just a set) for an abelian (non abelian) group.

Question

Let $G$ be an abelian Group.

Question is to prove that $T(G)=\{g\in G : |g|<\infty \}$ is a subgroup of G.

I tried in following way:

let $g_1,g_2\in T(G)$ say, $|g_1|=n_1$ and $|g_2|=n_2$;

Now, $(g_1g_2)^{n_1n_2}=g_1^{n_1n_2}g_2^{n_1n_2}$ [This is because G is abelian].

$(g_1g_2)^{n_1n_2}=g_1^{n_1n_2}g_2^{n_1n_2}=(g_1^{n_1})^{n_2}(g_2^{n_2})^{n_1}=e^{n_2}e^{n_1}=e$

Thus, if $g_1,g_2$ have finite order, so is $g_1g_2$.So, $T(G)$ is closed under group operation.

As $|g|< \infty$, suppose $|g|=n$ then, $g^n=e=g.g^{n-1}$

So, if we can see that $g^{n-1}$ is in $T(G)$, then we are done as $g^{n-1}$ would be inverse of $g$ in $T(G)$.

Now, $(g^{n-1})^n=(g^n)^{n-1}=e^{n-1}=e$.

So, $g^{n-1}$ is in $T(G)$ and thus we are done.[we did not use abelian property of G in proving existence of inverse]

So, we have $T(G)$ which is closed under group operation and inverse. Thus $T(G)$ is subgroup of $G$.

As i have not used abelianness (Sorry for this word :D) in one of the properties, Natural Question would be

Is $T(G)=\{g\in G : |g|<\infty \}$ a subgroup of G for non abelian G.

Only Non abelian Infinite Group that comes to my mind is $Gl_n(\mathbb{R})$ for a fixed $n\in \mathbb{N}$

It does not look so obvious for me to say $|A|<\infty, |B|<\infty$ implies $|AB|<\infty$,

I am not able to find an (an easy) example $A,B\in Gl_n(\mathbb{R})$ with $|A|<\infty, |B|<\infty$ but, $|AB|$ is not finite.

I am looking for an example (as requested above) and if possible another example of a nonabelian group of infinite order in which $T(G)$ would be seen to be not a subgroup with less effort/or atleast which you feel anybody should know.

Thanks in advance,

Regards, Praphulla Koushik

Answer 1

An example that immediately comes to mind is the infinite dihedral group. As motions of the plane this is generated by $s_1$ = reflection w.r.t. the line $x=0$, and by $s_2$ = reflection w.r.t. the line $x=1$. The composition $s_1\circ s_2$ is translation (in the direction of negative $x$-axis) by two units, and thus has infinite order in spite of the two reflections obviously being of order 2.

Answer 2

An example: $G=\langle a,b\mid a^2=b^2=1\rangle$. In it $|ab|=\infty$.

Answer 3

Matrix example:

Let $n=2$, $A=\begin{bmatrix} -1 & 1 \\ 0 & 1 \end{bmatrix}$, $B=\begin{bmatrix}-1&0\\0&1\end{bmatrix}$. Then $A$ and $B$ have order 2, but $AB = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ has infinite order.

This is just the matrix version of the infinite dihedral group mentioned in the other two answers. Matrices of the form $\begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix}$ with $a \neq 0$ form the affine general linear group $\operatorname{AGL}(1,\mathbb{R}) \leq \operatorname{GL}(2,\mathbb{R})$ and represent the affine functions $x \mapsto ax+b$. One replaces the number $x$ with the column vector $(x,1)$ then $\begin{bmatrix} a & b \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} x \\ 1 \end{bmatrix} = \begin{bmatrix} ax+b \\ 1 \end{bmatrix}$. At any rate, the infinite dihedral group acts on the real line and is generated by two reflectionss, $x \mapsto -x$ represented by $B$ and $x \mapsto 1-x$ represented by $A$.

Answer 4

You may see the answer through the group $\textbf{GL}(2,\mathbb Q)$ and the following members:

$$A=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},~~B=\begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}$$

It can easily be found out that $A^4=E=B^3$ but $|AB|=\infty$.

Source: Rotman's Theory of Group page 27.