Past exam of the short options course of the University of Oxford / 2015 / 2

The defining relation $yx=x^{4}y$ lets us move every occurrence of $y$ to the right of every occurrence of $x$. More generally, repeated conjugation givesNegative powers cause no problem because $x^{-1}=x^8$ and $y^{-1}=y^2$. Thus every word in the group presentation can be put in the required form, reducing powers using $x^9=y^3=e$:At this stage existence is proved; uniqueness will follow from the concrete semidirect product in part (b). Moving $x^r$ past $y^j$ gives the multiplication lawThe action is consistent with reduction of $j$ modulo $3$ because $4^3\equiv 1(\mathrm{mod}9)$.

For the given group homomorphism $\phi :H\to \mathrm{Aut}(N)$, the external semidirect product has underlying Cartesian product $N\times H$ and operationThe notation $\mathrm{Aut}(N)$ denotes the automorphism group of $N$. Closure follows because $\phi (h)$ is an automorphism of $N$. To check associativity, calculate both bracketings:They agree because $\phi (hk)=\phi (h)\phi (k)$. The identity is $(e_N,e_H)$, using $\phi (e_H)={\mathrm{id}}_N$, and the two-sided inverse isIndeed multiplication in either order gives the identity, using $\phi (h)\phi (h^{-1})={\mathrm{id}}_N$. This verifies the group axioms.

For the specified group presentation, take $N=C_9=⟨X⟩$ and $H=C_3=⟨Y⟩$. The rule $\alpha (X)=X^4$ defines an automorphism because $4$ is coprime to $9$. Furthermore $4^3\equiv 1(\mathrm{mod}9)$, so $Y\mapsto \alpha$ defines the required group homomorphism $C_3\to \mathrm{Aut}(C_9)$. In the resulting semidirect product $S$, the canonical generators satisfyConsequently the group presentation defines a surjective group homomorphism $G\to S$, sending $x$ to $X$ and $y$ to $Y$. The underlying Cartesian product gives $|S|=27$, whereas part (a) gives $|G|\le 27$. Surjectivity forces equality, so the map is an isomorphism:This also proves that the $27$ normal forms in part (a) are distinct and that $x$ has element order exactly $9$.

The cyclic subgroup $⟨x^3⟩$ is normal: conjugation by $y$ sends $x^3$ to $x^{12}=x^3$, and conjugation by $x$ fixes it. Modulo this subgroup, the relation $yx{y}^{-1}=x^4$ becomes $yx{y}^{-1}=x$, so the quotient group is Abelian. Thus the commutator subgroup $G^{\prime }$ is contained in $⟨x^3⟩$.

Conversely the displayed relation gives the commutator $yx{y}^{-1}{x}^{-1}=x^3$, so $⟨x^3⟩\subseteq G^{\prime }$. HenceSince $x$ has element order $9$, $x^3\ne e$, so the first commutator subgroup is nontrivial. ThereforeHere derived length counts the number of steps needed for the derived series to reach the trivial group.

The given unitriangular group of degree three over F3 has $27$ elements, just as $G$ does, so counting alone cannot distinguish them. Instead compare element orders.

Write an element as $I+U$, with $U$ strictly upper triangular of size $3$. Then $U^3=0$. Over the finite field ${\mathbb{F}}_3$, the binomial theorem for commuting matrices givesEvery nonidentity element of $H$ consequently has element order $3$. In contrast, $x\in G$ has element order $9$, by part (b). An isomorphism preserves element order, soEquivalently, the exponent of a finite group is $3$ for $H$ and $9$ for $G$.