Let \(G\) be a group, and \(Z(G)\) the center of \(G\). Prove that if \(G/Z(G)\) is cyclic, then \(G\) is abelian.
Prove that a group of order \(p^n\), where \(p\) is a prime and \(n \geq 1\), has non-trivial center.
Prove that a group of order \(p^2\) must be abelian.
Solution
1. A group with cyclic quotient by its center is abelian.
Choose \(a\in G\) whose coset generates \(G/Z(G)\). Every element of \(G\) has the form \(a^i z\) for some integer \(i\) and \(z\in Z(G)\). For any two such elements, centrality gives \[ (a^i z)(a^j w)=a^{i+j}zw =a^{j+i}wz=(a^j w)(a^i z). \] Thus every pair of elements commutes. The argument also covers the trivial cyclic quotient by taking \(a=1\).
2. A group of order \(p^n\), with \(n\geq1\), hasnontrivial center.
Let \(G\) act on itself by conjugation. The orbit of \(x\) has size \([G:C_G(x)]\) by orbit-stabilizer, where \(C_G(x)\) is its centralizer [@DF04]. The orbit has one element exactly when \(x\in Z(G)\). Every other orbit has size a power of \(p\) greater than one, and hence divisible by \(p\).
Partitioning \(G\) into its conjugacy classes therefore gives \(|G|\equiv|Z(G)|\pmod p\). Since \(p\mid|G|\), one has \(p\mid|Z(G)|\). The center contains the identity, so its positive order is at least \(p\).
3. Every group of order \(p^2\) is abelian.
By step 2 and Lagrange’s theorem, the center has order \(p\) or \(p^2\). In the second case \(Z(G)=G\), so the group is abelian. In the first case \(G/Z(G)\) has prime order \(p\). A group of prime order is cyclic: any nonidentity element generates a subgroup whose order divides \(p\) and exceeds one. Step 1 then again gives that \(G\) is abelian. In fact this rules out center order \(p\), leaving \(Z(G)=G\).