# Is $\{g \in G : |g| < \infty\}$ always subgroup of a group $G$? [duplicate]

Possible Duplicate:
$T(G)$ may not be a subgroup?

Let $G$ be a group, and consider $H = \{g \in G : |g| < \infty\}$.

Question: Must $H$ necessarily be a subgroup of $G$?

Here, $|g|$ denotes the order of the element $g$.

## marked as duplicate by Douglas S. Stones, martini, Sasha, Phira, Austin MohrNov 16 '12 at 7:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• What do the bars mean? Order? – martini Oct 22 '12 at 8:20
• This is not actually true unless you assume some more about $G$, for example it holds if $G$ is abelian. – Tobias Kildetoft Oct 22 '12 at 8:20
• This may be wrong, consider p. e. $\langle a,b \mid a^2 = b^2 = 1\rangle$, where $ab$ has infinite order... – martini Oct 22 '12 at 8:21
• I don't think this is true. What if $G$ is the group given by the presentation $\{ a, b ; a^2 = b^2 = 1 \}$. Then $|a| =|b|=2$ but $|ab|=\infty$. – MJD Oct 22 '12 at 8:23
• I edited the question to account for the observations after the original question was posted. – Douglas S. Stones Oct 23 '12 at 1:50

## 1 Answer

In general it is false that the subset of elements of a group $G$ of finite order is a subgroup. I think that the simplest, in some sense, case is that of $GL_2(\Bbb R)$. Let $s_1$ and $s_2$ be symmetries with respect to lines $\ell_1$ and $\ell_2$ through the origin. Then $s_1$ and $s_2$ have finite order (equal in fact to $2$) but the product $s_1s_2$ is a rotation whose order is finite if and only if the lines $\ell_1$ and $\ell_2$ form an angle which is a rational multiple of $2\pi$ (which is obviously not always the case).

However, the claim is true when the group $G$ is commutative. This follows immediately from the observation that if $ab=ba$ then the order of $ab$ divides the least common multiple of the orders of $a$ and $b$.