Does anybody know the answer to this or a good way to go about working this out? I have a list for $GL_2(Z/pZ)$ and I am trying to lift it to this; I have mostly been using fairly elementary algebraic methods. Thank you.

  • $\begingroup$ Steven Sam has a blog post about this at: concretenonsense.wordpress.com/2009/09/14/… $\endgroup$ – Gwyn Whieldon Mar 11 '14 at 13:44
  • 1
    $\begingroup$ Note that for this it is really important that it is not just the cyclic group of order $p^2$. It that group with the usual structure of a ring. $\endgroup$ – Tobias Kildetoft Mar 11 '14 at 13:58
  • $\begingroup$ @Tobias, I edited the title to reflect your comment. $\endgroup$ – Nick Gill Mar 11 '14 at 15:17
  • $\begingroup$ @Gwyn, I think Steven Sam's blogpost only covers conjugacy classes for $GL_2(q)$, i.e. over a finite field. It seems like the MO knows this theory, but is unclear how to lift to the ring $Z/p^2Z$.... I might be wrong! $\endgroup$ – Nick Gill Mar 11 '14 at 15:27
  • $\begingroup$ Ah, good point. I misread what he was asking for. $\endgroup$ – Gwyn Whieldon Mar 11 '14 at 15:33

I don't have time to give a full answer now, but here's how I would try and answer this question:

You first need to establish what the normal subgroups of $GL_2(\mathbb{Z}/p^2\mathbb{Z})$ are. This is done in a fair amount of detail at this MO question (at least for $SL_2$, and the same method will generalize to $GL_2$).

Now, as explained in the linked answer above, $G=GL_2(\mathbb{Z}/p^2\mathbb{Z})$ has a normal subgroup $N$ of order $p^3$ which can be thought of as a 3-dimensional module for the group $G/N\cong GL_2(\mathbb{Z}/p\mathbb{Z})$. To get the full-list of conjugacy classes, one needs to examine each coset $gN$ in turn, considering it as a module for the group $C_{G/N}(gN)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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