# Do these results also hold for automorphisms?

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.

• 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. Jan 26, 2020 at 1:08
• 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) Jan 26, 2020 at 10:26

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 $$2$$.
• 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 $Aut(C_2 \times C_2) \cong S_3$. Jan 26, 2020 at 10:21