TITLE: Aut($G$) $\simeq$ Aut($M$) $ \times$ Aut($N$)
QUESTION [6 upvotes]: A question in group theory:
Let $ G = M \times N $ be the direct product of $ 2 $ normal subgroups.
If $( | M | , | N | ) = 1 $ then Aut($G$) $\simeq$ Aut($M$) $ \times$ Aut($N$).
I proved that Aut($M$) $ \times$ Aut($N$) $\le$ Aut($G$), but I can't prove the other inclusion. Any hint ?

REPLY [2 votes]: Well Lagrange's theorem tells you you aren't going to have any $\phi\in\text{Aut}(G)$ move anything in $N$ to anything in $M$ and vice versa.  Thus we expect we can describe an automorphism in $\text{Aut}(G)$ by what it does to each of the direct factors - we want each $\phi$ to be written $\psi_M\times \psi_N$, where $(\psi_M\times\psi_N)(m,n)=(\psi_M(m),\psi_N(n))$.  Now, given a $\phi$, how could we figure out what $\psi_M$ and $\psi_N$ are?

 Consider $\phi(m,1)$.  Where does that map to? What about $\phi(1,n)$?  Remember that $M,N$ are normal and coprime, so they are characteristic Sylow subgroups of $M\times N$. Now what about $\phi(m,n)$?