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Let $G={\Bbb Z}_5\times {\Bbb Z}_{25}$. What is the order of $Aut(G)$ and is $Aut(G)$ abelian?


I don't know if this has anything to do with the result $$ Aut({\Bbb Z}_n)\cong U(n). $$

Does one have to count $Aut(G)$ by "brute force"? I've seen two similar questions:

But I don't see how this can be reduces to those two cases. I think that whether $Aut(G)$ is abelian can be partially told from its order.

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  • $\begingroup$ What is $U(n)?$ $\endgroup$ – Igor Rivin Jan 10 '14 at 19:09
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For abelian groups $G,H,K$ we have natural isomorphisms $$\hom(G,H \times K) \cong \hom(G,H) \times \hom(G,K),$$ $$\hom(G \times H,K) \cong \hom(G,K) \times \hom(H,K).$$ This enables you to compute $\hom(G,H)$ for arbitrary finite abelian groups $G,H$, as soon as you know that $\hom(\mathbb{Z}/n,\mathbb{Z}/m) \cong \mathbb{Z}/\mathrm{gcd}(n,m)$. It also follows that $\mathrm{End}(G)$ can be represented as a certain matrix ring, and $\mathrm{Aut}(G)$ is a group of invertible matrices. For the general procedure and all details, see

Christopher J. Hillar, Darren Rhea, Automorphisms of finite Abelian groups, arXiv

For example, if $p$ is a prime, then $$\mathrm{End}(\mathbb{Z}/p \times \mathbb{Z}/p^2) \cong \begin{pmatrix} \hom(\mathbb{Z}/p,\mathbb{Z}/p) & \hom(\mathbb{Z}/p^2,\mathbb{Z}/p) \\ \hom(\mathbb{Z}/p,\mathbb{Z}/p^2) & \hom(\mathbb{Z}/p^2,\mathbb{Z}/p^2) \end{pmatrix} \cong \begin{pmatrix} \mathbb{Z}/p & \mathbb{Z}/p \\ \mathbb{Z}/p & \mathbb{Z}/p^2 \end{pmatrix}$$

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  • $\begingroup$ Do you have a reference for the result in the first line? $\endgroup$ – Jack Jan 12 '14 at 2:12

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