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$T(G)$ 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 $g$.

  • 2
    $\begingroup$ What do the bars mean? Order? $\endgroup$
    – martini
    Oct 22 '12 at 8:20
  • 7
    $\begingroup$ This is not actually true unless you assume some more about $G$, for example it holds if $G$ is abelian. $\endgroup$ Oct 22 '12 at 8:20
  • 4
    $\begingroup$ This may be wrong, consider p. e. $\langle a,b \mid a^2 = b^2 = 1\rangle$, where $ab$ has infinite order... $\endgroup$
    – martini
    Oct 22 '12 at 8:21
  • 3
    $\begingroup$ 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 $|ab|=\infty$. $\endgroup$
    – MJD
    Oct 22 '12 at 8:23
  • $\begingroup$ I edited the question to account for the observations after the original question was posted. $\endgroup$ Oct 23 '12 at 1:50

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 $b$.


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