Let \(H\) be a finite \(p\)-group and let \(G\le H\). Suppose the composite \[ G\hookrightarrow H\twoheadrightarrow H/[H,H] \] is surjective. Prove that \(G=H\).
Solution
Surjectivity is equivalent to \[ H=G[H,H]. \] Suppose for contradiction that \(G<H\). Since \(H\) is finite, \(G\) is contained in a maximal subgroup \(M<H\).
1. Every maximal subgroup of a finite \(p\)-group has index \(p\) and is normal.
The quotient action of \(H\) on the left cosets \(H/M\) shows that the index is a power of \(p\). Maximality forces the quotient to have no nontrivial proper subgroup, hence \[ [H:M]=p. \] A subgroup of index \(p\) in a finite \(p\)-group is normal.
2. We have \([H,H]\subseteq M\).
Because \(H/M\) has order \(p\), it is cyclic and therefore abelian. The derived subgroup is contained in the kernel of every homomorphism to an abelian group, so \[ [H,H]\subseteq M. \]
Now \(G\subseteq M\) and \([H,H]\subseteq M\), so \[ G[H,H]\subseteq M<H, \] contradicting the surjectivity identity \(G[H,H]=H\). Therefore \(G=H\).