Let \(p\) be a prime and \(\abs{G} = p^3\). Prove that \(G\) has a normal subgroup \(N\) of order \(p^2\). - Suppose \(N = \gens{h}\) is cyclic and classify all possibilities for \(G\) if: - \(\abs h = p^3\) - \(\abs h = p\).
Hint: Sylow and semidirect products.
Hint
For a central \(z\) of order \(p\), the quotient \(G/\gens{z}\) has order \(p^2\) and is abelian; the preimage of a subgroup of order \(p\) is normal of order \(p^2\).
Solution
1. \(G\) has a normal subgroup of order \(p^2\).
1.1. \(G\) is a \(p\)-group, so the class equation\[\abs G = \abs{Z(G)} + \sum_{i} [G : C_G(x_i)]\]over representatives \(x_i\) of the noncentral conjugacy classes has every index \([G:C_G(x_i)]\) divisible by \(p\).
1.2. Hence \(p \mid \abs{Z(G)}\), so \(Z(G) \neq 1\).
1.3. By Cauchy’s theorem applied to \(Z(G)\), pick \(z \in Z(G)\) with \(\abs z = p\).
1.4. \(\gens z \normal G\), because a central subgroup is normalized by every element of \(G\).
1.5. \(\abs{G/\gens z} = p^3/p = p^2\), and every group of order \(p^2\) is abelian: if \(Q\) has order \(p^2\) then \(Q/Z(Q)\) is cyclic by step 2 applied to \(Q\), and a group with cyclic central quotient is abelian.
1.6. By Cauchy’s theorem, \(G/\gens z\) has a subgroup \(\bar H\) with \(\abs{\bar H} = p\), and \(\bar H \normal G/\gens z\) because that quotient is abelian.
1.7. Let \(N\) be the preimage of \(\bar H\) under \(\pi: G \to G/\gens z\). The correspondence theorem gives \(N \normal G\), and\[\abs N = \abs{\bar H}\cdot \abs{\gens z} = p \cdot p = p^2 .\]
2. The two classification cases are cases on the largest order of an element of \(G\).
2.1. \(N\) has order \(p^2\), so a generator of a cyclic \(N\) has order \(p^2\), never \(p^3\) or \(p\).
2.2. The orders \(p^3\) and \(p\) are therefore the orders of an element \(h \in G\) of largest order, and they are the two extreme cases: \(G\) is cyclic, or \(G\) has exponent \(p\).
3. If \(\abs h = p^3\) then \(G \cong \ZZ/p^3\).
3.1. \(\gens h \leq G\) has \(p^3 = \abs G\) elements, so \(\gens h = G\).
3.2. A cyclic group is determined up to isomorphism by its order.
4. If \(\abs h = p\) for an element of largest order, then \(G\) has exponent \(p\), and \(G\) is one of two groups.
4.1. Every nonidentity element of \(G\) has order \(p\), since \(p\) is the largest order occurring.
4.2. Suppose first that \(G\) is abelian. Then \(G\) is a vector space over \(\FF_p\) of dimension \(3\), so \(G \cong (\ZZ/p)^3\).
4.3. Suppose instead that \(G\) is nonabelian. Then \(\abs{Z(G)} \neq p^3\), and \(\abs{Z(G)} \neq p^2\), since \(G/Z(G)\) cyclic forces \(G\) abelian. With step <1>1’s \(Z(G) \neq 1\) this gives \(\abs{Z(G)} = p\).
4.4. So \(G/Z(G)\) has order \(p^2\) and is not cyclic, hence \(G/Z(G) \cong (\ZZ/p)^2\).
4.5. Choose \(x, y \in G\) whose images generate \(G/Z(G)\), and write \(Z(G) = \gens z\). Then \(G = \gens{x, y, z}\).
4.6. The commutator \([x,y]\) lies in \(Z(G)\), because \(G/Z(G)\) is abelian, and \([x,y] \neq 1\), because \(x\) and \(y\) do not commute. So \(\gens{[x,y]} = Z(G)\), and after replacing \(z\) by \([x,y]\) we may take\[x^p = y^p = z^p = 1, \qquad [x,y] = z, \qquad z \text{ central}.\]
4.7. These relations write every element of \(G\) once as \(x^a y^b z^c\) with \(0 \leq a,b,c < p\), so they present a single group of order \(p^3\), namely the group of upper unitriangular \(3\times 3\) matrices over \(\FF_p\),\[G \cong \ts{ \begin{pmatrix} 1 & a & c \\ 0 & 1 & b \\ 0 & 0 & 1\end{pmatrix} \st a,b,c \in \FF_p } \cong (\ZZ/p)^2 \semidirect \ZZ/p .\]
4.8. This case is empty for \(p = 2\): exponent \(2\) forces \(g^2 = 1\) for all \(g\), hence \((gh)^2 = 1\) and \(gh = hg\), so \(G\) is abelian and step 2 applies.
5. Q.E.D.
Step 1 produces \(N\), and steps 3 and 4 classify \(G\) in the two stated cases.