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Compute Aut$(\mathbb{Z}/p\mathbb{Z})^n$ up to isomorphism and its order when $p$ is a prime.

I have trouble understanding this homework exercise. Can someone explain what this exercise asks to compute?

user112358
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2 Answers2

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If you know a little linear algebra, you may be familiar with the concept of a vector space. In which consists of an abelian group being acted upon in a certain way by a field.

In this case, we have that $\mathbb{Z}/p\mathbb{Z}$ is a field and $(\mathbb{Z}/p\mathbb{Z})^{n}$ is a vector space over this field. Therefore, we can think of an automorphism $\alpha: (\mathbb{Z}/p\mathbb{Z})^{n} \rightarrow (\mathbb{Z}/p\mathbb{Z})^{n}$ as a linear transformation. Choosing a basis will give you an $n \times n$ matrix, and in order for this to be an automorphism, we must have that the matrix is invertible. This means that all of the columns must be linearly independent.

Here is a concrete example, consider $Aut(\mathbb{Z}/2\mathbb{Z})^{2})$. Our matrix will be of the form

$\begin{bmatrix} a &b \\ c&d \\ \end{bmatrix}$

for $a,b,c, d \in \mathbb{Z}/2\mathbb{Z}$. In order for the matrix to be invertible, so the first column must not be $[0,0]^{T}$. Therefore, we are left with three choices (4-1) for the first column. For the second column, we must not have it be a scalar multiple of the first column, this leaves us with two choices (4-2) for the second column. In total, we have $6=3 \times 2$ total elements.

user135520
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  • Thank you very much! This a very detailed answer. But now I am stuck when finding the order of $Aut(\mathbb{Z}/p\mathbb{Z})^n$ in general. Is there a systemmatic way to find how many elements are there? I find it hard to count the order when $p$ and $n$ are relatively large. Thank you! – user112358 Oct 30 '15 at 19:42
  • I'll get you started, you have $p^{n}-1$ possible entires for the first column. $p^{n}-p$ for the second column. $p^{n}-p^{2}$ entries for the third, and so forth. – user135520 Oct 30 '15 at 20:18
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Fir any group $G$, one can define the group $Aut(G)$ of group automorphisms of $G$. This the functions from $G$ to $G$, one-to-one, onto and that are group morphisms.

This is a group. The question is then, what is this group when $G=(\frac{\mathbb{Z}}{p\mathbb{Z}})^n$?

The answer should be a group with which you are more familiar (hint : It has something to do with matrices).

Clément Guérin
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