Let \(G\) be a group with normal subgroups \(N_1,N_2,N_3\). Suppose that for every \(i\ne j\), \[ N_i\cap N_j=\{e\}, \qquad N_iN_j=G. \] Show that \(G\) is abelian and that \(N_i\cong N_j\) for all \(i,j\).
Solution
1. If \(i\ne j\), then every element of \(N_i\) commutes with everyelement of \(N_j\).
Let \(x\in N_i\) and \(y\in N_j\). Since both subgroups are normal, \[ [x,y]=xyx^{-1}y^{-1} \] lies in \(N_i\) and in \(N_j\). Thus \[ [x,y]\in N_i\cap N_j=\{e\}, \] so \([x,y]=e\).
2. Each subgroup \(N_i\) is contained in the center \(Z(G)\).
Fix \(i\), and let \(j,k\) be the other two indices. By step 1, every element of \(N_i\) commutes with every element of both \(N_j\) and \(N_k\). The hypothesis gives \[ G=N_jN_k. \] Hence every element of \(G\) is a product \(yz\) with \[ y\in N_j, \qquad z\in N_k. \] An element of \(N_i\) commutes with both factors and therefore with their product. Thus \(N_i\subseteq Z(G)\).
3. The group \(G\) is abelian.
For example, \[ G=N_1N_2 \] by hypothesis. Step 2 gives \[ N_1,N_2\subseteq Z(G), \] so every element of \(G\) lies in \(Z(G)\). Hence \(G=Z(G)\) and \(G\) is abelian.
4. For every pair \(i\ne j\), the quotient map\[ q_i:G\longrightarrow G/N_i \]restricts to an isomorphism\[ q_i|_{N_j}:N_j\xrightarrow{\sim}G/N_i. \]
Its kernel is \[ N_j\cap N_i=\{e\}, \] so the restriction is injective. Its image is \[ q_i(N_j)=N_iN_j/N_i=G/N_i \] because \(N_iN_j=G\). Thus the restriction is also surjective.
5. The three subgroups are pairwise isomorphic:\[ \boxed{N_i\cong N_j\quad\text{for all }i,j.} \]
For any distinct indices \(i,j,k\), step 4 gives \[ N_i\cong G/N_k\cong N_j. \] The case \(i=j\) is immediate.
6. Q.E.D.
Step 3 proves that \(G\) is abelian, and step 5 proves the required pairwise isomorphisms.