TITLE: If $H$, $K$ are characteristic subgroups of $G$, when is $\operatorname{Aut}(H\times K) = \operatorname{Aut}(H) \times \operatorname{Aut}(K)$ true?
QUESTION [5 upvotes]: If $H$, $K$ are characteristic subgroups of $G$, when is $\operatorname{Aut}(H\times K) = \operatorname{Aut}(H) \times \operatorname{Aut}(K)$ true?
What if $H$, $K$ are not characteristic subgroups? 
In general, under what conditions is it true that $\operatorname{Aut}(H\times K) = \operatorname{Aut}(H) \times \operatorname{Aut}(K)$, for subgroups $H$, $K$ of $G$?

REPLY [2 votes]: it's the first time i try to answer my question ! i don't know if this will make the question closed or what will happen ,but i have a trial to prove this statement and want to share it with you ,and i hope there is no logical mistakes in it .
to prove $Aut(H \times K) = Aut(H) \times Aut(K) $
1- i prove that $Aut(H \times K) \subseteq Aut(H) \times Aut(K) $
let $M \in Aut(H \times K)$ then $M=(M_1,M_2)$
$M_1(H) \times M_2(K) = (M_1,M_2)(H\times K)=M(H\times K)=H\times K $
so , $M=(M_1,M_2) \in Aut(H)\times Aut(K)$
so, $Aut(H \times K) \subset Aut(H) \times Aut(K) $

2- i prove that $Aut(H) \times Aut(K)  \subseteq Aut(H \times K) $
let $M\in Aut(H)\times Aut(K) so M=(M_1,M_2) where M_1 \in Aut(H)$ and $M_2 \in Aut(K) $
$M(H\times K)=(M_1,M_2)(H\times K)=(M_1(H)\times M_2(K))=(H,K)$
so$M\in Aut(H\times K)$
so , $Aut(H) \times Aut(K)  \subseteq Aut(H \times K) $ 
using 1,2 ,we conclude that ,$Aut(H \times K) = Aut(H) \times Aut(K) $ 
i think that this is the Baddest proof i have ever written ! i don't know if it even makes sense !