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?

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    $\begingroup$ That looks correct but the normality comes into effect when you multiply two elements of $N_1N_2$. Essentially, $n_1n_2h_1h_2=n_1(n_2h_1n_2^{-1})n_2h_2$. The piece up to $n_2^{-1}$ is in $N_1$ and the rest is in $N_2$. $\endgroup$ – John Douma May 16 '13 at 4:50
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    $\begingroup$ Or, you can use 3 to establish an isomorphism $g\to(n_1,n_2)$. This may be another approach to get the answer. $\endgroup$ – Mikasa May 16 '13 at 4:55
  • $\begingroup$ @user69810 I am assuming that $n_1,h_1\in N_1$, $n_2,h_2\in N_2$? I can see that if $(n_2h_1n_2^{-1})\in N_2$ then $n_2h_1h_2\in N_2$, but where do you go from there? $\endgroup$ – Ruvi Lecamwasam May 16 '13 at 23:40
  • $\begingroup$ @BabakS. Thanks, I see how it makes sense to look at $G$ as a set of ordered pairs since each element can be written as the product. $\endgroup$ – Ruvi Lecamwasam May 16 '13 at 23:41
  • $\begingroup$ @AndrewLedesma Since $n_2h_1h_2\in N_2$ and $n_1\in N_1$, $n_1n_2h_1h_2\in N_1N_2$. Thus, $N_1N_2$ is closed under multiplication. $\endgroup$ – John Douma May 17 '13 at 0:09

I answered this question (starting from another one and generalizing it) in my post here. I hope it helps.

  • $\begingroup$ Thanks for that, it seems to go along the same lines as what I thought up. $\endgroup$ – Ruvi Lecamwasam May 16 '13 at 23:41

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