Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 10

A group representation assigns each $g\in G$ an invertible linear map $D(g)$ on a fixed vector space, preserving multiplication:Choosing a basis turns those maps into matrices. A faithful representation is injective, equivalently its kernel is $\{e\}$, so distinct elements have distinct matrices.

Two group representations on spaces of the same dimension are called equivalent representations if one fixed invertible change of basis conjugates every representing matrix: $E(g)=S^{-1}D(g)S$ for all $g$. Choosing a different $S$ separately for each element does not establish equivalence.

The character of a representation isThe cyclic property of the trace makes it invariant under a change of basis and constant on conjugacy classes. Thus different characters at even one group element prove inequivalence.

Since $a^4=I$, multiplication adds exponents modulo four: $a^j{a}^k=a^{j+k\mathrm{mod}4}$. Hence the Cayley table of the cyclic group of order four, with the left factor indexing rows, isThe table is symmetric because the group is Abelian. Its row for each element permutes all four members, as cancellation requires.

The matrices are listed in the order $I,a,a^2,a^3$. In the first representation, $D(a)$ is the clockwise quarter-turn matrix and $D(a{)}^2=-I_2$. In the second, $E(a)=i{I}_2$, whose successive powers give the listed scalar matrices. Both therefore obey the group table and distinguish all four elements.

Taking traces gives the characters of a representation:In particular ${\chi }_D(a)=0$ but ${\chi }_E(a)=2i$. Since trace is invariant under similarity, the representations are inequivalent, even after allowing complex changes of basis for the real matrices of $D$.

One can also see the obstruction directly: conjugating the scalar matrix $i{I}_2$ leaves it unchanged, whereas $D(a)$ is not scalar. Over the complex numbers, $D(a)$ has eigenvalues $i,-i$, while $E(a)$ has $i,i$; those different multiplicities cannot be removed by changing basis.

Write the lower block of $T(a)$ as $B=\left(\begin{array}{cc}0& b\\ c& 0\end{array}\right)$. ThenA representation of $C_4$ must satisfy its defining relation $T(a{)}^4=I_3$, so $(bc{)}^2=1$. For real $b,c$, this means $bc=1$ or $bc=-1$; it also ensures invertibility.

If $bc=-1$, then $T(a{)}^2\ne I_3$ while $T(a{)}^4=I_3$. The image of the generator has order four, so its four powers are distinct and the representation is faithful. If $bc=1$, then $T(a{)}^2=I_3$ but $T(a)\ne I_3$, since the lower block has zero diagonal. The image has order two and the kernel is $\{I,a^2\}$. HenceFor other values there is no representation of $C_4$, rather than an additional unfaithful case; in particular $bc=0$ would make the assigned matrix singular. These conclusions are the faithfulness criterion for a block representation of the cyclic group of order four. There is no requirement that the proposed representation be orthogonal, so no separate conditions $|b|=|c|=1$ should be imposed.