Consider the following elements of $GL_3(\mathbb{R})$:
$$A = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}\quad,\quad B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}\;.$$
Let $N$ be the subgroup of $GL_3(\mathbb{R})$ generated by $A$ and $B$.
(i) Does $N$ have any elements of order 4? Justify your answers.
(ii) Let $H$ be the cyclic subgroup of $GL_3(\mathbb{R})$ generated by $A$. Is $H$ normal in $N$? Is $H$ normal in $GL_3(\mathbb{R})$? Justify your answers.
I've calculated that $|A| = 3, |B| = 2 $ and $|AB| = 2$. Also $|AB| \neq |A||B|$. Am I meant to use Lagrange's theorem for (i). If so, how?
As for (ii) I am not sure how to answer this at all.
Thank you
By counting elements, you can show that $|N|=6$, so it can't have any elements of order $4$, as $4\nmid 6$. Note that this doesn't involve realizing $N$ as $S_3$.
As you mention, $|A|=3$ and hence $|H|=3$. Since $\frac{6}{3}=2$, $H$ is normal in $N$. Let me know if you need help with the last part about $H\vartriangleleft GL_3(\mathbb{R})$, but you should try for yourself.