Let \(G_1,G_2,G_3\) be finite groups, each generated by its commutators. Let \[ A\le G_1\times G_2\times G_3 \] be a subgroup whose natural projections onto each of \[ G_1\times G_2,\qquad G_1\times G_3,\qquad G_2\times G_3 \] are surjective. Prove that \[ A=G_1\times G_2\times G_3. \]
Solution
1. For every commutator \([x,y]\) in \(G_1\),\[ ([x,y],1,1)\in A. \]
By surjectivity of the projection \[ A\longrightarrow G_1\times G_2, \] there is an element \[ u=(x,1,u_3)\in A. \] By surjectivity of the projection \[ A\longrightarrow G_1\times G_3, \] there is an element \[ v=(y,v_2,1)\in A. \] Since \(A\) is a subgroup, it contains their commutator. Commutators in a direct product are computed coordinatewise, so \[ \begin{aligned} [u,v] &= ([x,y],[1,v_2],[u_3,1])\\ &= ([x,y],1,1). \end{aligned} \]
2. One has\[ G_1\times\{1\}\times\{1\}\leq A. \]
By hypothesis, \(G_1\) is generated by its commutators. Step 1 shows that the pure first-coordinate copy in \(A\) contains every such commutator. Since \(A\) is a subgroup, it therefore contains every finite product of these elements and hence every element of \[ G_1\times\{1\}\times\{1\}. \]
3. One has\[ \{1\}\times G_2\times\{1\}\leq A. \]
Let \(x,y\in G_2\). Surjectivity onto \(G_1\times G_2\) gives \[ u=(1,x,u_3)\in A, \] and surjectivity onto \(G_2\times G_3\) gives \[ v=(v_1,y,1)\in A. \] Then \[ [u,v]=(1,[x,y],1). \] Because \(G_2\) is generated by its commutators, the same argument as in step 2 gives the claimed inclusion.
4. One has\[ \{1\}\times\{1\}\times G_3\leq A. \]
Let \(x,y\in G_3\). Surjectivity onto \(G_1\times G_3\) gives \[ u=(1,u_2,x)\in A, \] and surjectivity onto \(G_2\times G_3\) gives \[ v=(v_1,1,y)\in A. \] Then \[ [u,v]=(1,1,[x,y]). \] Since \(G_3\) is generated by its commutators, this implies the claimed inclusion.
5. Therefore\[ \boxed{A=G_1\times G_2\times G_3}. \]
By steps <1>2–<1>4, for arbitrary \[ (g_1,g_2,g_3)\in G_1\times G_2\times G_3, \] the three elements \[ (g_1,1,1), \qquad (1,g_2,1), \qquad (1,1,g_3) \] all belong to \(A\). Their product is \((g_1,g_2,g_3)\), so the whole direct product is contained in \(A\). The reverse inclusion is part of the hypothesis that \(A\) is a subgroup of that direct product.
6. Q.E.D.
Step 5 is the required equality.