TITLE: $N_1,N_2,N_3 \unlhd G, N_i\cap N_j =\{e\}, G = N_iN_j$. Want to show that $G$ is abelian, $N_i$ are isomorphic.
QUESTION [7 upvotes]: The following is a problem from the Berkeley Problems book. 

Let $G$ be a group with three normal subgroups $N_1$ , $N_2$, and
  $N_3$. Suppose $N_i \cap N_j = \{ e\}$ and $N_iN_j  = G$ for all
  $(i,j )$ with $i \ne j$ . Show that $G$ is abelian and $N_i$, is
  isomorphic to $N_j$ for all $i, j$ .

Attempt: 
I have not got very far. The only thing that comes to my mind is that as $|HK|=\frac{|H||K|}{|H \cap K|}$, $|N_1|=|N_2|=|N_3|=m$, and order of $|G|=m^2$. I have made other futile attempts of trying to show m prime, finding an isomorphism between the $N's$. I think the fact that the equation $N_i N_j=G$, could be seen as, left coset of $N_j$ with the multiplying elements restricted to only $N_i$ could provide me an answer, but I haven't got anywhere with this. 
Thanks in advance.

REPLY [4 votes]: Let $x_i \in N_i$ and $y_j \in N_j$ with $i \neq j$. Then $[x_i,y_j]=(x_iy_jx_i^{-1})y_j=x_i(y_jx_i^{-1}y_j^{-1})$ so $[x_i,y_j] \in N_i \cap N_j$ and $[x_i,y_j]=1$. Now let $x,y \in N_i$; because $G=N_jN_k$ for $i,j,k$ pairwise non-equal, $x=x_jx_k$ with $x_j \in N_j$ and $x_k \in N_k$; but $y$ commutes with $x_j$ and $x_k$ by the remark above, therefore $y$ commutes with $x$ and $G$ is abelian.
Then, you can show that $G/N_i \simeq N_j$ and $G/N_i \simeq N_k$ with $i,j,k$ pairwise non-equal to conclude that $N_i \simeq N_j \simeq N_k$.
In fact, $G$ doesn't need to be finite.