Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let p be a prime number, and let X be the n-th power of the cyclic group $\mathbb{Z}_p$. Let G denote the automorphism group of X. Suppose we want to describe G as a matrix with coefficients in $\mathbb{Z}_p$. Am I correct in saying that G is a n by n matrix with coefficients in $\mathbb{Z}_p$ such that det(G) $\not\equiv$ 0 (mod p) (i.e., invertible).

Now I am working on the following problem: "Determine the order of G by using the orbit-stabilizer theorem and the action of G on X". I was thinking that I need to find the size of each orbit. Then the order of G is given by the least common multiple of the sizes of each orbit. Is it correct to proceed like this or is there a simpler way? Thanks in advance

share|improve this question
1  
Your first is right. For the second one, it is a standard exercise to calculate the size of $GL_n(p)$ by just noticing that a matrix has nonzero determinant iff its rows are linearly independent. –  ougao Dec 13 '12 at 18:48

1 Answer 1

up vote 1 down vote accepted

Yes, you are one-hundred percent correct. Namely, $\text{Aut}(\mathbb{Z}_p^n)\cong\text{GL}_n(\mathbb{Z}_p)$. Now, for the second part let's let $G=\text{GL}_n(\mathbb{Z}_p)$, $X=\mathbb{Z}_p^n$, let $G$ act on $X$ in the usual way. Take some $v\in X$ not equal to zero. Begin then by noting that $Gv=X-\{0\}$. Let's see if we can find what the stabilizer of $v$ is. Well, imagine taking a matrix in $\text{stab}(v)$ with respect to some basis $\{v,x_1,\cdots,x_{n-1}\}$ containing $v$. Now, we know that if we take the first column of this matrix with respect to $v$ we have $(1,0,\cdots,0)$ since $v$ maps to itself. Now, the only requirement on the second column is that it doesn't break invertibility, which is to say that it is linearly independent from the first column. Now, this is equivalent to saying that $x_1\notin\text{span}_{\mathbb{Z}_p})(v)$. But, since $|\text{span}_{\mathbb{Z}_p}(v)|=p$ there are $p^n-p$ such choices for $x_1$. Now, similarly, the only condition on choosing $x_2$ is that we require invertibility, which means choosing $x_2$ to be linearly independent from $\{v,x_1\}$ which means choosing $x_2$ not in $\text{span}_{\mathbb{Z}_p}(v,x_1)$. Since $|\text{span}_{\mathbb{Z}_p}\{v,x_1\}|=p^n-p^2$ we see that there are $p^n-p^2$ choices for $x_2$. Continuing in this fashion we find that the number of ways to pick the $2,\cdots,n$ columns of a matrix in $\text{stab}(v)$ are $(p^n-p)\cdots(p^n-p^{n-1})$ and so $|\text{stab}(v)|=(p^n-p)\cdots(p^n-p^{n-1})$. Now, the orbit stabilizer theorem tells us

$$|G|=|Gv||\text{stab}{v}|=(p^n-1)(p^n-p)\cdots(p^n-p^{n-1})$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.