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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 $i, j$ .


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.

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2 Answers 2

up vote 3 down vote accepted

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.

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+1 In fact each of $N_i$ is the complement of another in $G$ and so they are isomorphic. –  Babak S. Feb 4 '13 at 13:51
Should the second last line read "$G/N_i\cong N_j$ and $G/N_i\cong N_k$ with $i, j, k$ pairwise non-equal"? (Basically, should the first $\neq$ be a $\cong$, and I think you meant for the $\leq$ to be a $\neq$ but then you also need to say $i\neq k$ so just saying "pairwise non-equal" is neater, no?) –  user1729 Feb 4 '13 at 14:04
@user1729: Thank you for your comment, I edited my answer. I hope it is now completely clear. –  Seirios Feb 4 '13 at 15:55

Do you know that $G$ is a finite group? I would start with the following observation:

$$N_i,N_j \unlhd G, \,\,N_i\cap N_j = \{e\},\,\, N_i N_j = G \, \Longrightarrow G \cong N_i \oplus N_j$$

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This is an insufficient hint! A better one would be to note that $n_in_j=n_jn_i$ with $n_k\in N_k$, $i\neq j$. This does always happen in a direct product, but just saying "direct product!" doesn't lead you onto that unless you know that it holds for direct products already... –  user1729 Feb 4 '13 at 13:57
@Sam: It has been mentioned in the problem that G is a finite group. –  ramanujan_dirac Feb 4 '13 at 14:22
@Sam: I made a typo in my above comment. It HASNT been mentioned that G is finite. –  ramanujan_dirac Feb 4 '13 at 14:35
@user1729: a hint is a hint and not a complete solution ;) –  Sam Feb 5 '13 at 0:15

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