Let \(p\) be a prime and \(G\) a group of order \(p^3\).
Prove that \(G\) has a normal subgroup of order \(p^2\).
Assume that \(G\) has a cyclic normal subgroup \(N\) of order \(p^2\) generated by some element \(n\). Let \(g\) be an element not in \(N\).
If the order of \(g\) is \(p^3\), classify the possible \(G\) up to isomorphism.
- If the order of \(g\) is \(p\), classify the possible \(G\) up to isomorphism.
(Incidentally, for odd \(p\) there exist groups of neither type, such as the group of \(3 \times 3\) upper triangular matrices over \(\mathbb{F}_p\) with 1’s on the diagonal. For \(p=2\), the quaternion group \(Q_8\) is an example of neither type.)
The qualification “for odd \(p\)” in the upper-unitriangular example is necessary. For odd \(p\), every strictly upper triangular \(3\times3\) matrix \(X\) satisfies \(X^3=0\), and in characteristic \(p\) one has \((I+X)^p=I\), so the unitriangular group has exponent \(p\) and contains no cyclic subgroup of order \(p^2\). For \(p=2\), that matrix group instead has elements of order \(4\) and is the dihedral group of order \(8\), which occurs in part (b)(ii).
Solution
1. The group \(G\) has a normal subgroup of order \(p^2\).
The class equation for a finite \(p\)-group shows that its center is nontrivial. Choose a central subgroup \[ C\le Z(G),\qquad |C|=p, \] using Cauchy’s theorem.
The quotient \(G/C\) has order \(p^2\), and every group of order \(p^2\) is abelian. Indeed, its center is nontrivial; if the center has order \(p^2\) there is nothing to prove, while if it has order \(p\), the quotient by the center is cyclic, which forces the group itself to be abelian.
By Cauchy’s theorem, \(G/C\) has a subgroup \(K/C\) of order \(p\). Because \(G/C\) is abelian, \(K/C\) is normal in \(G/C\). Its inverse image \(K\) is therefore normal in \(G\), and \[ |K|=|C|\,|K/C|=p^2. \] This proves part (a).
2. In part (b)(i), the only possibility is \(G\cong C_{p^3}\).
If \(|g|=p^3\), then the cyclic subgroup \(\langle g\rangle\) already has \[ |\langle g\rangle|=p^3=|G|. \] Hence \(G=\langle g\rangle\) and \[ G\cong C_{p^3}. \] Conversely, the cyclic group \(C_{p^3}\) has its unique subgroup of order \(p^2\) as a cyclic normal subgroup, so this possibility occurs.
3. In part (b)(ii), \(G\) is a semidirect product \(C_{p^2}\rtimes C_p\).
Now assume \(|g|=p\). Since \(g\notin N\), the order-\(p\) subgroup \(\langle g\rangle\) is not contained in \(N\), and therefore \[ N\cap\langle g\rangle=1. \] Because \(N\trianglelefteq G\), the product \(N\langle g\rangle\) is a subgroup, and \[ |N\langle g\rangle| =\frac{|N|\,|\langle g\rangle|}{|N\cap\langle g\rangle|} =p^3. \] Thus \[ G=N\rtimes\langle g\rangle\cong C_{p^2}\rtimes C_p. \]
Write \(N=\langle n\rangle\). Conjugation by \(g\) is determined by a unit \(u\in(\mathbb Z/p^2\mathbb Z)^\times\) through \[ gng^{-1}=n^u. \] Since \(g^p=1\), the \(p\)th power of this automorphism is the identity, so \[ u^p\equiv1\pmod{p^2}. \]
4. If \(p\) is odd, part (b)(ii) gives exactly\[ C_{p^2}\times C_p \quad\text{and}\quad C_{p^2}\rtimes C_p, \]where in the nontrivial product a generator of \(C_p\) acts by \(n\mapsto n^{1+p}\).
Assume \(p\) is odd. Reducing \(u^p\equiv1\pmod{p^2}\) modulo \(p\) and using \(u^p\equiv u\pmod p\) gives \[ u\equiv1\pmod p. \] Hence \[ u\equiv1+kp\pmod{p^2} \] for a unique \(k\in\mathbb Z/p\mathbb Z\).
If \(k=0\), the action is trivial, so \[ G\cong C_{p^2}\times C_p. \] If \(k\ne0\), replace the generator \(g\) of \(\langle g\rangle\) by \(g^a\), where \[ ak\equiv1\pmod p. \] Then \[ (1+kp)^a\equiv1+akp\equiv1+p\pmod{p^2}, \] so every nontrivial action yields the same isomorphism type, with presentation \[ \left\langle n,g\ \middle|\ n^{p^2}=g^p=1,\ gng^{-1}=n^{1+p}\right\rangle. \] The two groups are not isomorphic because the first is abelian and the second is not.
5. If \(p=2\), part (b)(ii) gives exactly \(C_4\times C_2\) and the dihedral group of order \(8\).
For \(p=2\), \[ \operatorname{Aut}(C_4)\cong(\mathbb Z/4\mathbb Z)^\times=\{1,3\}. \] The trivial action gives \[ C_4\times C_2. \] The unique nontrivial action is inversion, \[ gng^{-1}=n^{-1}, \] and the resulting presentation \[ \langle n,g\mid n^4=g^2=1,\ gng^{-1}=n^{-1}\rangle \] is the dihedral group of order \(8\). Thus these are precisely the possibilities in part (b)(ii).