# Question on Abelian Group

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.

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What is a surefire technique to prove that a subset of a group is a subgroup? Have you tried applying such in this instance? Show some work. – anon Nov 15 '12 at 18:21
@anon: This is a duplicate. See math.stackexchange.com/q/208312/8581. – Babak S. Nov 15 '12 at 18:35
@Babak It certainly is a duplicate, but the OP may need to draw the connection between $|x| < \infty \implies$ that there is some integer $n$ such that $x^n = 0$, and may not have encountered the definition that all such elements are torsion elements. – amWhy Nov 15 '12 at 19:16
@amWhy: Ok. You are right. :) – Babak S. Nov 18 '12 at 17:56

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 $a$.

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There is another counter example which I think you may find it interesting. Please see Exercise 2.17 of J.J.Rotman's well-known book. There; he gave us $G=GL(2,\mathbb Q)$ such that $tG$ is not a subgroup..

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I assume that by $\lvert g \lvert$ you mean the order of $g$, so that the subset you are considering is the subset of all elements with finite order. Call this subset $X$.

Without actually doing the problem for you, here are a few things that might help.

To check that a subset of a group is a subgroup, all you need to do is to check that the subset itself satisfies the axioms for a group. First, then, you need to figure out whether if $g,h\in X$ then $gh\in X$. In your case this translates to the question about whether the product of two elements with finite order has finite order. Since your group $G$ is abelian, then this shouldn't be to hard to prove.

Then you need to check that all elements in $X$ have an inverse in $X$. An element $g\in G$ obviously has an inverse in $G$, but is this inverse actually in $X$? Well, if $g$ has finite orderm it shouldn't be too hard to see that $-g$ (writing the group additively because it is abelian) also has finite order (Hint: $-0 = 0$).

Without listing all the axioms, you also have to check that the addition of to elements of finite order has finite order.

Googeling or looking in a book will give you other things that you need to check.

As for the second question about an example of $G$ non-abelian where $X$ is not a subgroup, you can take a look at this qusetion: $T(G)$ may not be a subgroup? (As mentioned in the comments by @anon). Note, though, that in this question the group is written multiplicative.