Past exam of the short options course of the University of Oxford / 2021 / 2 (Group theory)

Let $|G|=p^2$. The allowed center of a group theorem gives $|Z(G)|>1$. By Lagrange's theorem its order is $p$ or $p^2$. In the first case $G/Z(G)$ has order $p$ and is a cyclic group. A cyclic quotient by the center forces $G$ to be Abelian: if $G/Z(G)=⟨gZ(G)⟩$, write arbitrary elements as $g^{a}z,g^{b}w$ with $z,w\in Z(G)$, and commute them directly. Hence in every caseFor two examples at prime-cube order, take the dihedral group of a square, $⟨r,s\mid r^4=s^2=1,srs=r^{-1}⟩$, and the quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$. Both have order $8=2^3$. They are nonabelian because $sr\ne rs$ and $ij=k\ne -k=ji$, respectively. They are distinct examples even up to group isomorphism: the dihedral group has five elements of order two and the quaternion group has only one. The request asks for two examples; it does not require the prime to be arbitrary.

Write $|G|=pq$ with $p<q$. The Sylow theorems give $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. Since $p<q$, necessarily $n_q=1$. Its unique Sylow subgroup $N$ is a normal subgroup and has prime order $q$, soA Sylow subgroup $H\cong C_p$ has trivial intersection with $N$, and $|NH|=|N||H|=pq$, so $NH=G$. ThusThe semidirect product action is a group homomorphism $C_p\to \mathrm{Aut}(C_q)\cong C_{q-1}$. If the action is trivial, $G\cong C_{pq}$ by the Chinese remainder theorem; if it is nontrivial, $p\mid q-1$. In particular $G$ has both a cyclic normal subgroup and a cyclic quotient, making it a metacyclic group and hence a solvable group.

The Abelian group criterion for a product of two prime squares applies because $11<19$ and $11\nmid 19^2-1=360$. Here its Sylow theorems proof is especially short:The candidates for $n_{19}$ are $1,11,121$, of which only $1$ has the required remainder. The candidates for $n_{11}$ are $1,19,361$, again leaving only $1$. Thus both Sylow subgroups $P,Q$ are normal subgroups. Their group commutators lie in $P\cap Q=\{1\}$, and $PQ=G$, so $G\cong P\times Q$. Both factors are abelian groups of prime-square order by (a).

The Fundamental theorem of finitely generated abelian groups gives preciselyThese four groups are pairwise nonisomorphic: their characteristic Sylow subgroups differ in whether they contain an element of order $11^2$ or $19^2$. The first is also the cyclic group $C_{43681}$.

Apply the Abelian group criterion for a product of two prime squares with the smaller and larger primes in their correct order. For $p<11$, all such groups are Abelian exactly when $p\nmid 11^2-1=120$. Of the primes $2,3,5,7$, only $7$ qualifies. For $p>11$, the criterion is $11\nmid p^2-1$, equivalently $p\not\equiv \pm 1(\mathrm{mod}11)$. Therefore the exact answer isEquivalently, among the stipulated primes $p\ne 11$, exclude $2,3,5$ and the residue classes $\pm 1$ modulo $11$.

For completeness, the necessity in that criterion supplies actual counterexamples at every excluded prime. If $\ell <q$ and $\ell \mid q^2-1$, the multiplicative group of a finite field is cyclic, so choose $\alpha \in {\mathbb{F}}_{q^2}^{\times }$ of order $\ell$. Let a generator of $C_{{\ell }^2}$ act on the additive group of ${\mathbb{F}}_{q^2}$ by multiplication by $\alpha$. This defines a nontrivial group action, hence a nonabelian semidirect product of order ${\ell }^2{q}^2$. Use $(\ell ,q)=(p,11)$ for $p=2,3,5$, and $(\ell ,q)=(11,p)$ for $p\equiv \pm 1(\mathrm{mod}11)$ with $p>11$. Thus no excluded prime can be restored to the answer.

Take $p=5$. The unit $3$ modulo $121$ has order $5$, becauseLet a generator $b$ of $C_{25}$ act on $C_{121}=⟨a⟩$ by $a\mapsto a^3$. The action is well defined since its order $5$ divides $25$. The semidirect producthas order $121\cdot 25=11^2{5}^2$. Its construction as a semidirect product ensures that $a$ retains order $121$. Since $a^3\ne a$, the groups generated by $a$ and $b$ do not commute, so $G$ is nonabelian.