If \(p\) is a prime, \(k\) is a positive integer and \(G\) is a group of order \(p^k\), show that for each positive integer \(i \leq k\), \(G\) has a normal subgroup of order \(p^i\).
Solution
1. Every nontrivial finite \(p\)-group has a central subgroup of order \(p\).
In the class equation, each noncentral conjugacy class has size \([G:C_G(x)]\), a power of \(p\) greater than \(1\) [@DF04]. It follows that \(p\mid |Z(G)|\). Since the center contains the identity, it contains some \(z\ne1\). By Lagrange’s theorem, the order of \(z\) is \(p^a\) with \(a\geq1\). Then \(c=z^{p^{a-1}}\) is nonidentity and satisfies \(c^p=1\), so it has order \(p\). The subgroup \(C=\langle c\rangle\) lies in the center and is therefore normal in \(G\).
2. For each integer \(1\leq i\leq k\), a group \(G\) of order \(p^k\) has a normal subgroup of order \(p^i\).
We induct on \(k\). When \(k=1\), the only order is \(p\), and \(G\) itself is a normal subgroup of that order.
1. It supplies the subgroup of order \(p\).Let \(\rho:G\to G/C\) be the quotient map; its codomain has order\(p^{k-1}\).
For any \(i\) with \(2\leq i\leq k\), induction supplies a normal subgroup \(M\lhd G/C\) of order \(p^{i-1}\). Set \(N=\rho^{-1}(M)\). For \(g\in G\) and \(n\in N\), we have \[ \rho(gng^{-1})=\rho(g)\rho(n)\rho(g)^{-1}\in M, \] so \(N\) is normal in \(G\). The map \(N\to M\) is surjective with kernel \(C\). Its fibers are cosets of \(C\), each of size \(p\); hence \(|N|=p|M|=p^i\). This constructs the required subgroup for every \(i\) and completes the induction.