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