# Automorphism group of a finite group

I would like to ask if there exists an explicit description of $\mathrm{Aut}(G)$, the group of automorphisms of a finite group $G$, in particular, when $G$ is abelian. E.g., if $G = \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$, where $m$ is a positive integer, how can we describe $\mathrm{Aut}(G)$? Which relation do we have between it and $GL_m(\mathbb{Z})$? If $m$ is prime then $\mathrm{Aut}(G) \cong GL_m(\mathbb{Z})$, but what happens for $m$ general? E.g., if $m=4$, I find that the cardinality of $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z})$ is $8$, by seeing to where the generators $(0,1)$ and $(1,0)$ are sent. But I have the feeling that $GL_4(\mathbb{Z})$ should be contained in $\mathrm{Aut}(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z})$, so I should have a problem with the cardinalities.

More general, is it sufficient to see where a minimal set of generators is sent? Maybe someone could indicate me a paper where I could find a good description of such automorphism groups?

• See this paper msri.org/people/members/chillar/files/autabeliangrps.pdf which I think appeared in the American Math Monthly. – Lucia Oct 15 '13 at 22:56
• For general results, see Bidwell, J.N.S., Curran, M.J., and McCaughan, D. Automorphisms of direct products of finite groups, Arch. Math. (Basel) 86 (2006) no. 6, 481-489; and Bidwell, J.N.S. Automorphisms of direct products of finite groups II. Arch. Math. (Basel) 91 (2008) no. 2, 111-121. – Arturo Magidin Oct 16 '13 at 4:34
• There is something wrong when you write that $GL_4(\mathbb{Z})$ should lie in $Aut(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z})$'', because the first is an infinite group! I presume that what you really mean is something like $GL_2(\mathbb{Z}/2\mathbb{Z})$ should lie in $Aut(\mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z})$(?) – Nick Gill Oct 16 '13 at 9:12
• While I'm at it, your first sentence suggests that you might be interested in non-abelian groups also... If this is the case, then Burnside's result on automorphisms of $p$-groups is very useful. (See p.174 of Gorenstein's "Finite groups".) – Nick Gill Oct 16 '13 at 9:14
• @NickGill $GL_2(\mathbf Z/2\mathbf Z)$ is a quotient of $\mathrm{Aut}(\mathbf Z/4\mathbf Z\times \mathbf Z/4\mathbf Z)$. The kernel is an abelian group isomorphic to the additive group of $2\times 2$ matrices $M_2(\mathbf Z/2\mathbf Z)$. – Amritanshu Prasad Oct 17 '13 at 6:44

Note that each finite abelian group is the product of its $p$-primary subgroups. An automorphism preserves the primary parts. So it is sufficient to consider $p$-abelian groups.
It is convenient to think of automorphisms of finite abelian groups as integer matrices. Expressing the group $A = \mathbf Z/p^{\lambda_1}\oplus \dotsb \oplus \mathbf Z/p^{\lambda_n}$ as a quotient of the free abelian group $\mathbf Z^n$, lift an automorphism $\phi$ of $A$ to an automorphism $\tilde\phi$ of $\mathbf Z^n$: $$\begin{matrix} \mathbf Z^n & \xrightarrow{\tilde \phi} & \mathbf Z^n\\ \downarrow & &\downarrow\\ A & \xrightarrow{\phi} & A \end{matrix}$$ The matrix $(\phi_{ij})$ representing $\tilde\phi$ is an invertible integer matrix. As far as the automorphism $\phi$ is concerned, the its entries in the $i$th row are in $\mathbf Z/p^{\lambda_i}\mathbf Z$. Also $\phi_{ij}$ is divisible by $p^{\max(0, \lambda_j-\lambda_i)}$. With this matrix representation, it is easy to do calculations. For example, composition is matrix multiplication.
As for your query concerning cardinalities: if $\lambda_1>\lambda_2>\dotsb>\lambda_n$ and $$A = (\mathbf Z/p^{\lambda_1}\mathbf Z)^{\oplus m_1}\oplus\dotsc \oplus (\mathbf Z/p^{\lambda_n}\mathbf Z)^{\oplus m_n},$$ then it is possible to deduce from the above description of the automorphism group that $$\lvert\mathrm{Aut}(A)\rvert = q^{\sum_{i,j}m_im_j\min(\lambda_i,\lambda_j)}\prod_{k=1}^n \prod_{l=1}^{m_k} (1 - q^{-l}).$$ The group $\prod_{k=1}^n GL_{m_k}(\mathbf Z/p\mathbf Z)$ is a quotient of $\mathrm{Aut}(A)$ by a $p$-group.
• Just in case anyone else falls into this trap, it is not true that an automorphism of $A$ lifts to an automorphism of $\mathbb{Z}^n$. Simply take $A = \mathbb{Z}/5\mathbb{Z}$ and let $\phi : A \to A$ be the automorphism defined by $\phi(1+5\mathbb{Z}) = 2+5\mathbb{Z}$. This does not lift to an automorphism of $\mathbb{Z}$ (but it does lift to an endomorphism). A complete solution to when such an automorphism lifts is given in the paper "Tame Automorphisms of Finitely Generated Abelian Groups" by Turner and Voce. – Jay Taylor Jul 7 '17 at 6:43