Let \(G\) have normal subgroups \(N_1,N_2,N_3\) such that \(N_i\cap N_j=\{e\}\) and \(N_iN_j=G\) whenever \(i\ne j\). Show that \(G\) is abelian and the \(N_i\) are pairwise isomorphic.
Solution
1. Distinct \(N_i\) and \(N_j\) commute elementwise.
Let \(i\ne j\), and take \[ x\in N_i, \qquad y\in N_j. \] Because \(N_i\) is normal, \[ [x,y]=xyx^{-1}y^{-1}\in N_i. \] Because \(N_j\) is normal, the same commutator also belongs to \(N_j\). Hence \[ [x,y]\in N_i\cap N_j=\{e\}. \] Thus \(xy=yx\).
2. Identify \(G\) with the direct product \(N_1\times N_2\).
By hypothesis, \[ G=N_1N_2. \] Step 1 shows that \(N_1\) and \(N_2\) commute elementwise, and their intersection is trivial. Therefore the multiplication map \[ N_1\times N_2\longrightarrow G, \qquad (x,y)\longmapsto xy, \] is an isomorphism. We henceforth use this identification.
3. The subgroup \(N_3\) is the graph of an isomorphism \(\varphi:N_1\to N_2\).
Let \[ \pi_1:G=N_1\times N_2\to N_1 \] be the first projection. Its restriction to \(N_3\) has kernel \[ N_3\cap N_2=\{e\}, \] so it is injective. Also \[ N_3N_2=G, \] so every first coordinate occurs among elements of \(N_3\); hence \(\pi_1|_{N_3}\) is surjective. Thus it is an isomorphism.
Similarly, the second projection restricts to an isomorphism \[ \pi_2|_{N_3}:N_3\to N_2. \] Define \[ \varphi =(\pi_2|_{N_3})\circ(\pi_1|_{N_3})^{-1}:N_1\to N_2. \] Then \(\varphi\) is an isomorphism and \[ N_3=\{(x,\varphi(x)):x\in N_1\}. \] In particular \[ N_1\cong N_3\cong N_2. \]
4. Normality of \(N_3\) forces \(N_1\) to be abelian.
Take \(a,x\in N_1\). Since \[ (x,\varphi(x))\in N_3 \] and \(N_3\trianglelefteq G\), conjugating by \((a,e)\) gives \[ (a,e)(x,\varphi(x))(a,e)^{-1} =(axa^{-1},\varphi(x))\in N_3. \] But \(N_3\) is the graph of \(\varphi\), so membership in \(N_3\) implies \[ \varphi(axa^{-1})=\varphi(x). \] Since \(\varphi\) is injective, \[ axa^{-1}=x. \] Thus every \(a,x\in N_1\) commute, and \(N_1\) is abelian. Since \(N_2\cong N_1\) and \(N_3\cong N_1\), all three are abelian.
5. Conclude that \(G\) is abelian.
By step 2, \[ G\cong N_1\times N_2. \] Both factors are abelian by step 4, so their direct product is abelian. Therefore \[ \boxed{G\text{ is abelian}, \qquad N_1\cong N_2\cong N_3.} \]