I am trying to get some ideas on the following problem but no result. Please show me the way.

Aut$(G)$ denotes the group of automorphisms of the group $G$. If $p, q$ are distinct primes then how shall we find the isomorphism class of $$Aut(\mathbb{Z}_{p^a}\times \mathbb{Z}_{p^b q^c})$$ where $a, b, c\in \mathbb{N}$.

What I tried is the following: $\mathbb{Z}_{p^bq^c}\simeq \mathbb{Z}_{p^b}\times \mathbb{Z}_{q^c}$ and hence $$Aut(\mathbb{Z}_{p^a}\times \mathbb{Z}_{p^b q^c})\simeq Aut(\mathbb{Z}_{p^a}\times \mathbb{Z}_{p^b})\times Aut(\mathbb{Z}_{q^c})$$ because we know that if $(|G_1|, |G_2|)=1$ then $Aut(G_1\times G_2)\simeq Aut(G_1)\times Aut(G_2)$.

Question is: What shall we do now? Is there any new ideas please or am I completely in the wrong direction ?

Thanks in advance

  • $\begingroup$ A good step would be figuring out what $\mathbb{Z}_{p^a}\times\mathbb{Z}_{p^b}$ is isomoprhic too :) $\endgroup$ – Zelos Malum Dec 2 '15 at 6:19

It is well-known that Aut($\mathbb{Z}_{q^c}$) has order $q^{c-1}(q-1)$, and it's cyclic except in the case $q=2$ and $c \ge 3$, when it's the product of two cyclic groups one of which has order 2. So the only question is, what is Aut($\mathbb{Z}_{p^a} \times \mathbb{Z}_{p^b}$)?

In the case $a=b$, it is not too hard to see that the answer is $GL(2,\mathbb{Z}_{p^a})$ (invertible 2x2 matrices over the ring $\mathbb{Z}_{p^a}$).

In the general case, I will assume $a \le b$. The automorphism group turns out to be a group of "2x2 matrices with mixed entries": $\begin{pmatrix}A & B\\C & D\end{pmatrix}$ where $A, B \in \mathbb{Z}_{p^a}$ ; $C, D \in \mathbb{Z}_{p^b}$; determinant not divisible by $p$; and $C \equiv 0 \mod{p^{b-a}}$.

These matrices act as automorphisms on $\mathbb{Z}_{p^a} \times \mathbb{Z}_{p^b}$ (as column vectors) in the usual way, and the matrices multiply by each other in the usual way. You might be worried that you're adding/multiplying elements from different rings, but it turns out everything is well-defined. (Lift everything to $\mathbb{Z}$, do the arithmetic in $\mathbb{Z}$, and project back down; then check that the result is independent of the lift.)

The condition $C \equiv 0 \mod{p^{b-a}}$ is important for things to work out. You can see that this condition is necessary because the image of $(1,0)$ is $(A,C)$, and the former has order $p^a$ so the latter must also, thus forcing $C \equiv 0 \mod{p^{b-a}}$.

For a precise treatment of the above, see this article on the group of automorphisms of an arbitrary finite abelian group.

  • $\begingroup$ nice and beautiful answer indeed. I was looking for such a reply only. Thank you once again $\endgroup$ – Anjan3 Dec 2 '15 at 7:07

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