Let \(G\) be a non-abelian group of order \(p^3\) with \(p\) a prime.
Determine the order of the center \(Z\) of \(G\).
Determine the number of inequivalent complex 1-dimensional representations of \(G\).
Compute the dimensions of all the inequivalent irreducible representations of \(G\) and verify that the number of such representations equals the number of conjugacy classes of \(G\).
Solution
1. The center \(Z=Z(G)\) has order \(p\).
Every nontrivial finite \(p\)-group has nontrivial center, so \(|Z|\ge p\). Since \(G\) is nonabelian, \(Z\ne G\). If \(|Z|=p^2\), then \(G/Z\) has order \(p\), hence is cyclic; but if \(G/Z(G)\) is cyclic, then \(G\) is abelian, a contradiction. Therefore \[ |Z|=p. \]
2. We have\[ G/Z\cong C_p\times C_p \quad\text{and}\quad [G,G]=Z. \]
The quotient \(G/Z\) has order \(p^2\). It cannot be cyclic, since that would force \(G\) abelian. Hence \[ G/Z\cong C_p\times C_p. \] In particular \(G/Z\) is abelian, so \[ [G,G]\subseteq Z. \] Because \(G\) is nonabelian, \([G,G]\ne1\). Since \(|Z|=p\), it follows that \[ [G,G]=Z. \]
3. The number of inequivalent one-dimensional complex representations of \(G\) is \(p^2\).
One-dimensional complex representations factor through the abelianization \[ G^{\mathrm{ab}}=G/[G,G]. \] By 2, \[ |G^{\mathrm{ab}}|=|G/Z|=p^2. \] A finite abelian group \(A\) has exactly \(|A|\) complex linear characters. Therefore \(G\) has exactly \[ p^2 \] inequivalent one-dimensional representations.
4. Every noncentral element of \(G\) has centralizer of order \(p^2\), and hence conjugacy class of size \(p\).
Let \(x\notin Z\). Its centralizer \(C_G(x)\) contains both \(x\) and \(Z\), so it contains the subgroup \(\langle x,Z\rangle\). Since \(x\notin Z\), this subgroup has order at least \(p^2\). On the other hand \(C_G(x)\ne G\), because \(x\) is not central. Thus \[ |C_G(x)|=p^2. \] By the orbit-stabilizer formula for conjugation, \[ |x^G|=[G:C_G(x)]=p. \]
5. The number of conjugacy classes of \(G\) is\[ p^2+p-1. \]
The \(p\) elements of \(Z\) each form a singleton conjugacy class. The remaining \[ p^3-p \] elements are partitioned, by 4, into conjugacy classes of size \(p\). Hence the number of noncentral conjugacy classes is \[ \frac{p^3-p}{p}=p^2-1. \] Thus the total number is \[ p+(p^2-1)=p^2+p-1. \]
6. Therefore \(G\) has exactly \(p-1\) nonlinear irreducible complex representations.
For a finite group, the number of inequivalent irreducible complex representations equals the number of conjugacy classes. By 5 there are \(p^2+p-1\) irreducibles in total, and by 3 exactly \(p^2\) are one-dimensional. Hence the number of nonlinear irreducibles is \[ (p^2+p-1)-p^2=p-1. \]
7. Every irreducible complex representation of \(G\) has dimension at most \(p\).
Let \(\chi\) be an irreducible character of degree \(d=\chi(1)\). By Schur’s lemma, every \(z\in Z\) acts by a scalar of absolute value \(1\), so \[ |\chi(z)|=d \qquad(z\in Z). \] Character orthogonality gives \[ \sum_{g\in G}|\chi(g)|^2=|G|=p^3. \] Restricting the sum to the \(p\) central elements yields \[ pd^2\le p^3, \] so \[ d\le p. \]
8. Every nonlinear irreducible representation has dimension exactly \(p\).
Let the nonlinear irreducible degrees be \[ d_1,\dots,d_{p-1}. \] The sum-of-squares formula gives \[ |G|=\sum_{\rho\in\operatorname{Irr}(G)}(\dim\rho)^2. \] The \(p^2\) linear representations contribute \(p^2\), so \[ \sum_{j=1}^{p-1}d_j^2=p^3-p^2=p^2(p-1). \] By 7, each \(d_j^2\le p^2\). Since there are exactly \(p-1\) terms and their sum is exactly \(p^2(p-1)\), equality must hold term-by-term: \[ d_j=p \qquad(1\le j\le p-1). \]
9. Thus the irreducible representation degrees are\[ \boxed{\underbrace{1,\dots,1}_{p^2\text{ times}}, \underbrace{p,\dots,p}_{p-1\text{ times}}}, \]and their number is\[ p^2+p-1, \]which agrees with the number of conjugacy classes computed in
5.