# Group does not contain any elements of order $p^2$?

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!

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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.
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.