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.

How should I Describe automorphism groups $\operatorname{Aut}(\mathbb{Z}_9)$ and $\operatorname{Aut}(\mathbb{Z}_3 \times \mathbb{Z}_3)$ ?

If a group G contains a normal subgroup H of order 9, G is generated by H and an element x not in H of order 3, How should I classify all such groups G?

share|improve this question
1  
Can you elaborate on what you consider to be a description? –  Austin Mohr Dec 13 '12 at 0:24

1 Answer 1

Ok, let's take the two separately.

Begin by noting that $\text{Aut}(\mathbb{Z}_9)\to \mathbb{Z}_n^\times$ (where the latter is the unit group) given by $f\mapsto f(1)$ is an isomorphism. So, you know that $\text{Aut}(\mathbb{Z}_9)\cong \mathbb{Z}_9^\times$. Now, it is a common fact the invertible elements of $\mathbb{Z}_n$ are those coprime to $n$. Thus, you at least know that $|\mathbb{Z}_9^\times|=\varphi(9)=6$. Since there is only one abelian group of order $6$ you may conclude that $\mathbb{Z}_9^\times\cong\mathbb{Z}_6$.

Now, the case of $\text{Aut}(\mathbb{Z}_3^2)$ is a little bit tricker. The key is to notice that $\mathbb{Z}_3^2$ is just a two-dimensional $\mathbb{Z}_3$-vector space and that group maps are really just $\mathbb{Z}_3$-linear maps. So, the invertible group maps $\mathbb{Z}_3^2\to\mathbb{Z}_3^2$ are precisely the invertible $\mathbb{Z}_3$-linear maps $\mathbb{Z}_3^2\to\mathbb{Z}_3^2$. But, this last set of maps has a nice description. Namely, by picking a basis for $\mathbb{Z}_3^2$, say $\{(1,0),(0,1)\}$ you can realize isomorphically the set of $\mathbb{Z}_3$-linear maps as just the $2\times2$ invertible matrices over $\mathbb{Z}_3$. Thus, you see that, up to isomorphism, $\text{Aut}(\mathbb{Z}_3^2)$ is just $\text{GL}_2(\mathbb{Z}_3)$.

Now, both of these results have generalizations. The second one is easier. Namely, using the exact same logic, I bet you can prove that for any prime $p$ and any integer $n$ that $\text{Aut}(\mathbb{Z}_p^n)\cong\text{GL}_n(\mathbb{Z}_p)$.

The first is a little harder. You can actually prove, using the exact same technique I mentioned, that $\text{Aut}(\mathbb{Z}_n)$ ($n$ some integer) is just $(\mathbb{Z}_n)^\times$. Now, to find $\mathbb{Z}_n^\times$ it suffices to find $\mathbb{Z}_{p^m}^\times$ for $p$ a prime. This is because, as rings, $\mathbb{Z}_n$ splits into a product of rings of the form $\mathbb{Z}_{p^m}$ (by the Chinese Remainder Theorem). Since this is a ring isomorphism, the unit groups of the two are isomorphic and since the unit group of a product is the product group of the units you get that $\mathbb{Z}_n^\times$ is just a product of groups of the form $\mathbb{Z}_{p^m}^\times$. Now, to find these is a little bit tricky, but you do get the following result:

$$\mathbb{Z}_{p^m}^\times\cong\begin{cases}\mathbb{Z}_{p^{m-1}(p-1)} & \mbox{if}\quad p\text{ is odd}\\ \{e\} & \mbox{if}\quad p=2, m=1\\ \mathbb{Z}_2 & \mbox{if}\quad p=2,m=2\\ \mathbb{Z}_{2^{m-2}}\times\mathbb{Z}_2 & \mbox{if}\quad p=2,m\geqslant 3 \end{cases}$$

If you are really interested in the proof, you can see my blog post about it here.

share|improve this answer
    
If a group G contains a normal subgroup H of order 9, G is generated by H and an element x not in H of order 3, How should I classify all such groups G? –  tom Dec 13 '12 at 5:36
    
@tom: Ask you second question in a new thread. –  user1729 Dec 13 '12 at 11:28

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.