1

Suppose $G$ is a finite abelian $p$-group. How to determine the order of $\operatorname{Aut}(G)$?

If $G$ has the form $\mathbb Z_{p^e}^n$, then the problem is not too difficult; the automorphism group is isomorphic to $\operatorname{GL}_n(\mathbb Z_{p^e})$, and its order is given by $(p^{en}-p^{en-n})(p^{en}-p^{en-n+1})\cdots(p^{en}-p^{en-1})$ according to my computation. But what if $G$ doesn't have the form? Is there any general method to compute $|\operatorname{Aut}(G)|$ in this case? For a concrete example, what is the order of $\operatorname{Aut}(\mathbb Z_2 \oplus\mathbb Z_2 \oplus \mathbb Z_4)$?

Any help is appreciated, including solving the example $\mathbb Z_2 \oplus\mathbb Z_2 \oplus \mathbb Z_4$.

EDIT: Many thanks to both references, which have fully solved my problem. Thanks.

Cyankite
  • 1,049
  • See https://arxiv.org/abs/math/0605185 – Steve D Apr 24 '24 at 12:26
  • The marked duplicate reduces the problem to counting certain matrices; in your case, the entries of the matrix correspond to maps $C_k\to C_m$, and these are well understood and correspond to the integers $r$ modulo $m$ such that $r/\gcd(m,r)$ is not relatively prime to $k$. Computing the number of such matrices which are invertible is fairly straightforward, so that question gives all the information required to answer your question. (cont) – Arturo Magidin Apr 25 '24 at 04:09
  • (cont) For instance, for $C_2\oplus C_2\oplus C_4$, the automorphisms correspond to $3\times 3$ matrices where the $(1,1)$, $(2,2)$, $(1,2)$, $(2,1)$, $(3,1)$, and $(3,2)$ entries are integers modulo $2$; the $(3,3)$ entry is an integer modulo $4$, and the remaining entries are either $0$ or $2$, taken modulo $4$. Then you just need to count how many of those matrices are invertible. Which is what the article does. The linked to question is an "abstract duplicate" in that it provides all the necessary information even if it does not provide an already-digested explicit answer. – Arturo Magidin Apr 25 '24 at 04:16
  • @ArturoMagidin I understand now. You are absolutely right. I have deleted the last few words in my post that disagree on marking duplicate, and my submission can be rejected:) – Cyankite Apr 25 '24 at 05:21

0 Answers0