Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to show that the group of $3 \times 3 $ upper triangular matrices over the field $ \mathbb{F}_p $ with diagonal entries 1 does not contain any elements of order $p^2$ when $ p \geq 3$.

I've tried to argue by contradiction: suppose it had an element $g$ of order $p^2$, then the subgroup generated by $g$ has index $p$ so it is normal, and from there I would like to find an element $h$ of order $p$ whose cyclic subgroup has trivial intersection with $ \langle g \rangle$. Then $G$ is the semi-direct product of $\langle g \rangle$ and $ \langle h \rangle$, and I am hoping this will give me a contradiction by telling me that $G$ is abelian or something similar.

Any help is appreciated!

share|cite|improve this question
up vote 3 down vote accepted

To show that the group does not have any elements of order $p^2$, it suffices to show that every element has order at most $p$. Equivalently, it suffices to show that $x^p = 1$ for every element $x$. This can be done by direct calculation. Calculate $A^2, A^3, A^4, A^5, \ldots$ for $$A = \left( \begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix} \right)$$

and you should see a pattern. Once you have a formula for $A^k$, the rest should be easy.

share|cite|improve this answer
though it may look a bit messy at first glance in the upper right corner – Hagen von Eitzen Nov 18 '12 at 18:27
Thanks for the help! I did attempt to prove $x^p = 1$ by direct computation earlier, but did not go to high enough powers to see the pattern. – Jonas Nov 18 '12 at 19:24

The matrix $$A = \left( \begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix} \right)$$ has minimum polynomial $(x-1)^3$ so in particular satisfies the polynomial $(x-1)^p = x^p - 1$ as long as $p \geq 3$. For $p=2$ (or fields of characteristic 2), when $a=c \neq 0$, then the matrix has order 4, not 2. For fields of characteristic 0, the group has no non-identity elements of finite order.

share|cite|improve this answer
Very nice way of looking at the problem, thanks! – Jonas Nov 19 '12 at 0:11

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.