Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 5 (Unitriangular groups and cubing)

In the finite field ${\mathbb{Z}}_3=\{0,1,2\}$, perform ordinary addition and multiplication and then take the remainder modulo $3$. The tables areThe nonzero elements have multiplicative inverses $1^{-1}=1$ and $2^{-1}=2$.

Parametrize the matrices byDirect matrix multiplication givesThis proves closure. The identity matrix is $h(0,0,0)$, and the formulagives both a left and a right inverse, as substitution in the product formula verifies. Thus each element is an invertible matrix, and the subgroup test shows that $H$ is a subgroup of ${\mathrm{GL}}_3({\mathbb{Z}}_3)$. It is the unitriangular group ${\mathrm{UT}}_3({\mathbb{F}}_3)$ and has $3^3=27$ elements.

The identity is the identity matrixLeft or right matrix multiplication by it preserves every entry, as is also immediate from the product formula above.

Take $x=h(1,0,0)=I_3+E_{12}$ and $y=h(0,0,1)=I_3+E_{23}$. Their matrix products areThey differ in the $(1,3)$ entry because $E_{12}{E}_{23}=E_{13}$ whereas $E_{23}{E}_{12}=0$. Therefore $xy\ne yx$, and $\boxed{H\text{ is a non-abelian group}}$.

If $G$ is an abelian group, its elements commute, so for all $x,y\in G$,Thus $\boxed{T\text{ is a group homomorphism}}$. Also $T(e)=e^3=e$, as required.

Use the symmetric group $S_3$, composing permutations from right to left. Set $x=(12)$ and $y=(23)$. Each is a transposition, so $x^3=x$ and $y^3=y$, whereas $xy=(123)$ has order of a group element three. ConsequentlyThus the cubing map on $S_3$ is not a group homomorphism.

Assume $T$ is a group homomorphism. For any $a,b\in G$, expanding $(ab{)}^3=a^3{b}^3$ and cancelling the initial $a$ and final $b$ gives the useful identityWe also need a consequence of compatibility with conjugation. Calculate the cube of $gz{g}^{-1}$ in two ways:Cancelling shows $g^2{z}^3=z^3{g}^2$. Thus the square of every element of $G$ commutes with every element of the image of a group homomorphism $J=T(G)$.

Now take any $x,y\in J$. In particular $x^{2}y=y{x}^2$. Applying (1) to $x,y$ and then using this commutation givesThis proves the abelian image of a cubing homomorphism property. Since the image of a group homomorphism is a subgroup, $\boxed{T(G)\text{ is an abelian group}}$.

If $T$ is a surjective group homomorphism, then $G=T(G)$, which is an abelian group by the preceding part.

If $T$ is an injective group homomorphism, then for all $a,b\in G$,because $T(a),T(b)$ lie in the abelian group $T(G)$. Injectivity gives $ab=ba$. Hence $\boxed{G\text{ is abelian in either case}}$.

Yes: use the non-abelian group $H$ from part (a). Every element is $I_3+N$ for a strictly upper triangular matrix $N$ with $N^3=0$. In the finite field ${\mathbb{Z}}_3$, the binomial theorem for commuting matrices givesThus $\boxed{T(h)=I_3\text{ for all }h\in H}$. This constant identity-valued map is a group homomorphism, since $T(xy)=I_3=T(x)T(y)$. Part (a)(iv) already proves that $H$ is not an abelian group. Here $T$ is neither injective nor surjective, consistently with the preceding part.