Let $G$ be an abelian group.
Show that $\{x\in{G} | |x| < \infty\}$ is a subgroup of $G$. Give an example of a non-abelian group where this fails to be a subgroup.
[Math] Elements of finite order in an abelian group form a subgroup
abstract-algebragroup-theory
Best Answer
Hint: Here is one of many ways of constructing an example. Let $G$ be the group of permutations of the integers. Let $f$ be the permutation that takes any integer $x$ to $-x$, and $g$ the permutation that takes any integer $x$ to $1-x$.
Both $f$ and $g$ have order $2$. Now consider the permutation $gf$, meaning $f$, followed by $g$. Show that $gf$ does not have finite order.
If you prefer matrices, let $A=\begin{pmatrix}-1 &0\\0&1\end{pmatrix}$ and $B=\begin{pmatrix}-1 &1\\0&1\end{pmatrix}$.
Then $A^2$ and $B^2$ are both the identity matrix. But $BA$ has infinite order. To see this, check what $BA$ does to the vector $\begin{pmatrix}n \\1\end{pmatrix}$
Remark: For the Abelian case, we need to show closure under product and inverse. For product, note that if $a^m=e$ and $b^n=e$, then $(ab)^{mn}=a^{mn}b^{mn}=e$. Inverse is easier, since in any group, Abelian or not, the inverse of $a$ has the same order as $a$.