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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 $g$.

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    I edited the question to account for the observations after the original question was posted.2012-10-23

1 Answers 1

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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 $b$.