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I would like help with the following problem:

Let $G$ be the subgroup of $GL_3(\mathbb{C})$ generated by the three matrices $ A=\begin{pmatrix} 0&0&1\\0&1&0\\1&0&0\end{pmatrix}\;,\quad B=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}\;,\quad C=\begin{pmatrix} i&0&0\\0&1&0\\0&0&1\end{pmatrix}$ 1. Compute the order of $G$. 2. Find a matrix $G$ of largest possible order (as an element of $G$) and compute this order. 3. Compute the number of elements in $G$ with this largest order.

So I've found the relations $A^2=B^3=C^4=I$, so I know my group so far consists of $\{I,A,B,B^2,C,C^2,C^3\}$, but I don't know how to proceed from here. I know the group is not abelian, but I just thought I'd try listing all two element products, so I have 30=6(5) choices for that (assuming, which I haven't checked, they are all distinct). I am just wondering if there is a more concrete way to approach this problem. I'd also like help with the other parts afterward. I can see that $BC$ and $CB$ have order 12, but I'm not sure what to do next.

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    @ Qiaochu Yuan, is this also obvious so you can shed light on it: https://math.stackexchange.com/questions/2731225/2018-04-10

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The subgroup generated by $A, B$ is of order 6. You can see this by checking that $ABA=B^{-1}$. Then look at $ACA^{-1}$, $BCB^{-1}$ this gives two more diagonal matrices which generate a subgroup of 64 diagonal matrices which is normalized by the group generated by $A,B$. So the order is 384.

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    Reading a word in the group generated by $A, B$, from left to right, when you get to an $AB$ in the word replace it with $B^{-1}A$, then all the $Bs$ accumulate on the left...2012-08-22
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This is an extended hint, not a full solution, and it takes advantage of some special features of the problem that won't be available in general. (On the other hand, this problem is quite hard in general, so one expects to have to use special features of any particular case anyway.)

Your matrices $A, B, C$ are all contained in the group of $3 \times 3$ generalized permutation matrices whose nonzero entries are allowed to be in $\{ 1, -1, i, -i \}$. This is the wreath product $C_4 \wr S_3$, and by counting the possible permutations and then the possible entries we see that it has order $6 \cdot 4^3 = 384$. One way to solve this problem is to figure out what subgroup of this wreath product you get using $A, B, C$.

Alternately, figure out what group you get using just $A$ and $B$, and then figure out how this group acts by conjugation on $C$. This will allow you to express $G$ as a certain semidirect product.