Let G be a non-abelian finite group. Show that \(G / Z ( G )\) is not cyclic, where \(Z ( G )\) is the center of G.
If \(| G | = p ^ { n }\) , with p prime and \(n > 0\) , show that \(Z ( G )\) is not trivial.
If \(| G | = p ^ { 2 }\) , show that G is abelian.
Solution
1. If \(G/Z(G)\) is cyclic, then \(G\) is abelian.
Suppose \[ G/Z(G)=\langle gZ(G)\rangle. \] For arbitrary \(x,y\in G\), there exist integers \(r,s\) and central elements \(z_1,z_2\in Z(G)\) such that \[ x=g^rz_1, \qquad y=g^sz_2. \] Since \(z_1\) and \(z_2\) commute with every element of \(G\), \[ xy = g^{r+s}z_1z_2 = g^{r+s}z_2z_1 = yx. \] Thus every pair of elements of \(G\) commutes.
2. If \(G\) is nonabelian, then \(G/Z(G)\) is not cyclic.
This is the contrapositive of step 1, and proves part 1.
3. If \(\abs G=p^n\) with \(n>0\), then\[ p\mid\abs{Z(G)}. \]
For \(x\in G\), the conjugacy class of \(x\) has cardinality \[ [G:C_G(x)]. \] This index divides \(\abs G=p^n\), so every conjugacy-class size is a power of \(p\). The class has size \(1\) exactly when \(x\in Z(G)\). Hence every noncentral conjugacy class has size divisible by \(p\).
The class equation therefore has the form \[ \abs G = \abs{Z(G)} + \sum_j p^{r_j}, \qquad r_j\geq1. \] Reducing modulo \(p\) and using \(p\mid\abs G\) gives \[ \abs{Z(G)}\equiv0\pmod p. \]
4. If \(\abs G=p^n\) with \(n>0\), then \(Z(G)\) is nontrivial.
The identity element lies in \(Z(G)\), so \(\abs{Z(G)}>0\). By step 3, this positive integer is divisible by \(p\), hence \[ \abs{Z(G)}\geq p>1. \] This proves part 2.
5. If \(\abs G=p^2\), then \(G\) is abelian.
By step 4, the center has order divisible by \(p\). Since its order also divides \(p^2\), either \[ \abs{Z(G)}=p \] or \[ \abs{Z(G)}=p^2. \] In the second case, \(Z(G)=G\), so \(G\) is abelian.
In the first case, \[ \abs{G/Z(G)}=p. \] Every group of prime order is cyclic, so \(G/Z(G)\) is cyclic. Step 1 then implies that \(G\) is abelian. This proves part 3.
6. Q.E.D.
Step 2 proves part 1, step 4 proves part 2, and step 5 proves part 3.