Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 1 / 19I (Representation Theory)

If $g$ belongs to the centre of a group, then $\rho (g)$ commutes with $\rho (h)$ for every $h\in G$. The Schur lemma says that an endomorphism of a finite-dimensional complex irreducible representation commuting with the representation is scalar. ThereforeSince $g$ has finite order, ${\lambda }_g$ is a root of unity and has modulus one.

Let $\chi$ be the character of a representation of $\rho$. The character orthogonality identity for an irreducible character is ${\sum }_{g\in G}|\chi (g){|}^2=|G|=n$. For $g\in Z$, the scalar formula from part (i) gives $|\chi (g)|=d$. Keeping these $m$ nonnegative summands yields

If $\rho$ is a faithful representation, $g\mapsto {\lambda }_g$ embeds $Z$ into the multiplicative group of complex numbers. Its image is a finite group of roots of unity, hence lies in the cyclic group of $n$th roots of unity and is itself cyclic. Thus $Z$ is cyclic.

For the order-18 group, the index-two subgroup $P$ is normal and $G=⟨P,t⟩$. Restrict a complex irreducible representation $V$ to the finite abelian group $P$. Simultaneous diagonalization decomposes it into character spaces $V_{\chi }$. The inversion relation makes $t$ exchange $V_{\chi }$ and $V_{{\chi }^{-1}}$. The sum of spaces belonging to any such character orbit is $G$-invariant, so irreducibility leaves only one orbit.

If the character is trivial, all of $P$ is in the representation's kernel. Otherwise $\chi$ has order three and $\chi \ne {\chi }^{-1}$. The only spaces are $V_{\chi }$ and $V_{{\chi }^{-1}}$, and both have the same kernel in $P$. A nontrivial character of $P\cong C_3\times C_3$ maps onto $C_3$, so that kernel has order three. It acts trivially on all of $V$. Therefore the group has no faithful irreducible complex representation. In fact choosing nonzero $v\in V_{\chi }$ makes $\mathrm{span}(v,tv)$ invariant, so these nontrivial-character irreducibles have degree two.