Give an example of a non-abelian group $G$ such that $H = \{a \in G \mid\operatorname{ord}(a) < \infty\}$ is not a subgroup of $G$.

Question

Give and example of a non-abelian group $G$ such that $H = \{a \in G\mid\operatorname{ord}(a) < \infty\}$ is not a subgroup of $G$.

Firstly, I thought about $GL_2(\mathbb{R})$, as multiplication of matrices is not commutative, I tried to shrink $GL_2(\mathbb{R})$ by introducing some special cases, as for example $G = \left\{\begin{pmatrix} 1 & a \\ 0 & b \end{pmatrix} : b \neq 0\right\}$ and found $H = \left\{\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & a \\ 0 & -1 \end{pmatrix} : a\in \mathbb R \right\}$, but in fact H is subgroup of G. Any suggestions?

Take $G = GL_2(\mathbb R)$, and assume $H = \{a \in G \mid\operatorname{ord}(a) < \infty\}$ is subgroup of $G$. Consider $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1/2 \\ 2 & 0 \end{pmatrix} \in H$, as $ \ \ \ \ \ \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = I$ and $\begin{pmatrix} 0 & 1/2 \\ 2 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1/2 \\ 2 & 0 \end{pmatrix} = I$

$W = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0 & 1/2 \\ 2 & 0 \end{pmatrix} = \begin{pmatrix} 2 & 0 \\ 0 & 1/2 \end{pmatrix} \not \in H$ as $ord(W) = \infty$, therefore by Subgroup Criterion contradiction $\implies$ $H$ is not subgroup of $G$.

Answer 1

Maybe think about the dihedral analogue but on a circle instead of a polygon. Pick a reflection about a line of symmetry through 1 radian and one about 2 radians (you just need them to not be a multiple of $\pi$). Then each is its own inverse so they would be in this torsion set since their order is $2$. But when you multiply them you should get just a rotation of an integer multiple of radians. This should have infinite order because it will never get to $2\pi n$ since if it did, it would mean $\pi$ is rational.