Compute Aut$(\mathbb{Z}/p\mathbb{Z})^n$ up to isomorphism 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?
 A: 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.
A: 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).
