I have come across the statement:
$G$ is the direct product of its subgroups $N_1$ and $N_2$ if the following conditions hold:
1) $N_1,N_2$ are normal subgroups.
2) $N_1\cap N_2=\{e\}$
3) They generate the group, i.e. $G=N_1N_2$
The book had previously defined the external direct product of two groups as consisting of ordered pairs of elements, one from each group. I felt that this wouldn't make much sense in this context, so I assumed this meant you could write each element of $G$ uniquely as a product of $n_1n_2$ where $n_1\in N_1,n_2\in N_2$. Then to prove this I wrote if $n_1n_2=n_1'n_2'$ then we have $n_1'^{-1}n_1=n_2'n_2^{-1}$, and since $N_1\cap N_2=\{e\}$ we have $n_1=n_1',n_2=n_2'$. This didn't however use the fact that these are normal subgroups, and I thought that maybe this is necessary because suppose $n_1n_2\ne e$, then $n_1n_2=n_2(n_2^{-1}n_1n_2)$. As $n_2\in N_2$ the product in the brackets must be in $N_1$, since $n_1n_2\ne e$ and the subsets are trivially disjoint.
I think this is correct, but the book doesn't say anything more on the subject so I can't be sure. Is this chain of reasoning correct?
Best Answer
I answered this question (starting from another one and generalizing it) in my post here. I hope it helps.