Is there a way to describe the structure of the automorphism group of $$C_{p}^2 \rtimes C_p \cong \langle x, y, z | [x,y]=z, [x,z]=[y,z]=x^p=y^p=z^p=e \rangle \cong UT(3, p)?$$ Here $p$ is an odd prime.

The only thing I know about it is, that $Inn(UT(3, p)) \cong \frac{UT(3, p)}{Z(UT(3, p))} \cong C_p \times C_p$, however $UT(3, p)$ is also very likely to have outer automorphisms, which I do not know how to describe.

Also, one can see, that all inner automorphisms of $UT(3, p)$ are of the form $$\begin{pmatrix} 1 & x & y \\ 0 & 1 & z\\ 0 & 0 & 1 \end{pmatrix} \mapsto \begin{pmatrix} 1 & x & y + (a-c)z -ac\\ 0 & 1 & z\\ 0 & 0 & 1 \end{pmatrix}$$ for some $a, c \in \mathbb{F}_p$. However, this does not help much.

  • $\begingroup$ This paper might help. $\endgroup$ – James Mar 14 at 22:04
  • $\begingroup$ I think you have a typo, the presentation is for $C_p^2\rtimes C_p$. $\endgroup$ – Sam Hughes Mar 23 at 22:39

For $p\geq3$ we have $Out(G)=GL_2(p)$ and $Aut(G)=AGL_2(p)\cong C_p^2\rtimes GL_2(p)$. To see this it helps to think of the group as $\mathbb{F}_p^2\rtimes C_p$ with the $C_p$ factor acting linearly, then the outer automorphisms come from $GL_2(p)$ acting by conjugation. Equivalently, the automorphisms essentially permute the subgroups of order $p^2$ containing the center, while leaving the center itself unmoved.

This may be useful as well.


As pointed out by Derek Holt in the comments, viewing the group as the above semidirect product is not the clearest way to see the automorphism group. It is better to think of $Out(G)$ as a sympletic group preserving the alternating form defined by commutators in $G$. This will still give $Out(G)\cong GL_2(p)$ and $Aut(G)\cong AGL_2(p)$.

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    $\begingroup$ I am not convinced that it is particularly helpful to think of the group as a semidirect product in the way you describe, because it can be written like that in many different ways, none of which is invariant under the automorphism group. I would think of Out($G$) as a symplectic group preserving the alternating form defined by commutators in $G$. $\endgroup$ – Derek Holt Mar 24 at 0:41
  • $\begingroup$ Yes you are right, that is a bit of an oversight on my part. Do you mind if I add your comment on to the answer? $\endgroup$ – Sam Hughes Mar 24 at 10:42
  • $\begingroup$ No, not at all. $\endgroup$ – Derek Holt Mar 24 at 10:42

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