The direct product of quotients is a quotient of the direct product

Question

I asked a question about if we have in general that if $G=G_1\times \cdots \times G_n$ (where $G_i$ are characteristic in $G$ for $i=1,\cdots ,n$), then $${\rm Out}(G)\cong {\rm Out}(G_1)\times\cdots\times {\rm Out}(G_n).$$

A comment suggested that I should use the following two facts:

1. The inner automorphism group of a direct product is the direct product of the inner automorphism groups.
2. The direct product of quotients is a quotient of the direct product.

I can prove the first one. I know how to prove a similar result for ${\rm Aut}(G)$. In addition, the ${\rm Aut}(G)$ case requires that those $G_i$ be characteristic, while the ${\rm Inn}(G)$ case only requires normality.

But how to prove the second one? Is it really true in general? Any help is appreciated.

Answer 1

It's true if $A$ and $B$ are normal in $G$ and $H$ respectively. Then $A×B\triangleleft G×H$. Define $i:(G×H)/(A×B)\to G/A×H/B$ by $i((g,h)+(A×B))=(g+A,h+B)$. It's straight forward to check that $i$ is an isomorphism.

Answer 2

HINT

If $H\lhd G$ and $H'\lhd G'$ then $f:G\times G'\to (G/H)\times (G'/H')$ with $f((x,x'))=(pr(x),pr'(x'))$ has kernel $H\times H'$