Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
Have you tried the Todd-Coxeter algorithm? Alternately, there is a not-terribly-complicated finite group which contains $G$ as a subgroup (permutation matrices with entries in $\{ 1, -1, i, -i \}$) and you could try to compute the index of $G$ in this group. –  Qiaochu Yuan Aug 20 '12 at 23:35
I don't know for sure that this will work, but you might try computing commutators like $A^{-1}B^{-1}AB$. If you can express them all in the form $A^iB^jC^k$ then you can get a standard form for elements of the group, and that will give you the order. Even if it doesn't quite work out that way you may get enough relations to get the kind of information you want. –  Gerry Myerson Aug 20 '12 at 23:37
I'm not familiar with that algorithm, but I can try learning it and maybe it will lead to a solution. Would you mind please elaborating more on the other method with the permutation matrices? –  mathmath8128 Aug 21 '12 at 0:38

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer
could you please explain how seeing $ABA=B^{-1}$ implies that the subgroup is of order 6? I think I could figure out the latter claims if I could have some help understanding the first part. Thank you. –  mathmath8128 Aug 21 '12 at 5:59
Since $ABA=B^{-1}$ then the subgroup of order 3 generated by $B$ is normal. Note also that $BA=B^{-1}A$. Then show every element in the group generated by $A, B$ can be written as $B^iA^j$ for for $i=0,1,2, j=0,1$. –  i. m. soloveichik Aug 21 '12 at 11:26
I get now that $ABA=B^{-1}$ implies the subgroup generated by $B$ is normal in the subgroup generated by $A,B$, but I still don't understand the latter claims. Would you mind explaining further? What can I conclude once I know the subgroup is normal? I'm not getting any insight on how to prove the subgroup generated by $A,B$ can be written like $B^iA^j$... I hope you can help further. –  mathmath8128 Aug 22 '12 at 10:23
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... –  i. m. soloveichik Aug 22 '12 at 11:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.