 Solution

A group homomorphism $D:G_1\to G_2$ satisfies $D(gh)=D(g)D(h)$ for all $g,h\in G_1$. Its group kernel is

For the matrices, put $X=R(x)$, $Y=R(y)$ and $z=e^{2\pi i/3}$. Since $z^3=1$, $z\ne 1$ and $1+z+z^2=0$, direct matrix multiplication givesThese verify the listed matrices and the defining relations, with $R(e)=I_2$. To check every product explicitly through a single calculation, write elements in the unique form $x^{ϵ}{y}^j$, with $ϵ=0,1$ and $j=0,1,2$. The relations giveThe displayed matrices obey $Y^{j}X=X{Y}^{-j}$, so their product has exactly the same form:All exponents are reduced modulo two or three respectively. Thus this verifies all 36 products, and $R$ is a two-dimensional complex group representation. It is given to be an irreducible representation.

For a one-dimensional complex group representation $U$, let $u=U(x)$ and $v=U(y)$, which are nonzero complex numbers. The relations requireSince scalar multiplication commutes and $u\ne 0$, the last relation gives $v=v^2$, hence $v=1$. Then $u=\pm 1$. The choice $u=1$ is the trivial representation; the unique nontrivial choice is the sign representation:Its group kernel isA faithful representation is injective and has trivial kernel. Here the kernel has three elements, so $U$ is not faithful.

Both one-dimensional representations are irreducible representations, and they are inequivalent to each other and to the given two-dimensional one. The sum of squares of irreducible representation dimensions givesThese exhaust all complex irreducible representations. Their number equals the number of conjugacy classes, so there are three conjugacy classes. Explicitly they areThe last two sets consist of the 3-cycles and transpositions respectively.