Let $G, G_1, G_2,..., G_n$ be groups.

I know the following results:

a) If $G$ is abelian, then $\operatorname{Hom}(G_1 \times G_2 \times \dots \times G_n, G)\cong \operatorname{Hom}(G_1,G)\times \operatorname{Hom}(G_2,G)\times \dots \times \operatorname{Hom}(G_n,G)$
(you may find a proof here: https://ysharifi.wordpress.com/2019/09/26/group-homomorphism-direct-product-1/ )

b) If $G_1, G_2,..., G_n$ are all abelian, then $\operatorname{Hom}(G, G_1 \times G_2 \times \dots \times G_n)\cong \operatorname{Hom}(G, G_1)\times \operatorname{Hom}(G, G_2) \times \dots \times \operatorname{Hom}(G, G_n)$
(you may find a proof here: https://ysharifi.wordpress.com/2019/09/26/group-homomorphism-direct-product-2/)

I was wondering if these results also hold if we replace $\operatorname{Hom}$ by $\operatorname{Aut}$. I believe that they should, since automorphisms are homomorphism, so the proofs from the links above should still work). If they do not, I would be interested if something similar exists for $\operatorname{Aut}$.

EDIT: As pointed out, these results do not hold for $\operatorname{Aut}$. I would like to know if there is something similar for $\operatorname{Aut}$, though.
A little bit more info: when dealing with the number of endomorphisms of an abelian group these results are pretty useful because I can use the structure theorem and then apply them. Now I am interested if something similar can be done for automorphisms.

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    $\begingroup$ For (a), the map from the product is surjective but not usually injective. For (b), it is the opposite: the map to the product is injective but not usually surjective. $\endgroup$ Commented Jan 26, 2020 at 1:08
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    $\begingroup$ If you take $C_2 \times C_2 \times C_2$ you see that its automorphism group is simple ant thus not decomposable into any non-trivial direct products: groupprops.subwiki.org/wiki/GL(3,2) $\endgroup$ Commented Jan 26, 2020 at 10:26

1 Answer 1


This is not true for $Aut$ since $Aut(\mathbb{Z}/2\times\mathbb{Z}/2)$ is distint of $Aut(\mathbb{Z}/2)\times Aut(\mathbb{Z}/2)$, the second group is embedded in the first group and does not contain the map $(x,y)\rightarrow (y,x)$ since the groups are finite they dont have the same cardinal.

Remark that if $f$ is an automorphism of $\mathbb{Z}/2$, $f(\bar 0)=\bar 0$ since $\bar 0$ is the neutral, $f(\bar 1)=\bar 1$ since $f$ is bijective, $f$ is the identity. We deduce that $Aut(\mathbb{Z}/2)\times Aut(\mathbb{Z}/2)$ contains only one element, but $Aut(\mathbb{Z}/2\times\mathbb{Z}/2)$ contains the identity and $f(x,y)=(y,x)$ its cardinality is at least $2$.

  • $\begingroup$ The cardinality of $Aut(C_2 \times C_2)$ is not only $\geq 2$ , it is even equal to $6$. That is because any permutation of the three non-identity elements of $C_2 \times C_2$ induces a unique automorphism (one can check the manually). Thus we have $Aut(C_2 \times C_2) \cong S_3$. $\endgroup$ Commented Jan 26, 2020 at 10:21

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