Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 19I (Representation Theory)

A representation is completely reducible, or semisimple, when it is a direct sum of irreducible representation. Equivalently, every invariant subspace has an invariant complement.

Maschke's theorem says that every finite-dimensional representation of a finite group $G$ over a field of characteristic zero is completely reducible. More generally, it is enough that the field characteristic does not divide $|G|$.

Schur lemma states the following.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility forcesso $T$ is an isomorphism. For an endomorphism of a finite-dimensional complex representation, choose an eigenvalue $\lambda$ of $T$. The intertwiner $T-\lambda I$ has nonzero kernel, so the first part forcesThis proves the Proof of Schur lemma.

Now let $\rho :G\to \mathrm{GL}(V)$ be faithful and irreducible over $\mathbb{C}$. For every $z\in Z(G)$, the operator $\rho (z)$ commutes with every $\rho (g)$. Schur's lemma therefore givesfor some ${\lambda }_z\in {\mathbb{C}}^{\times }$. Faithfulness makes $z\mapsto {\lambda }_z$ an embeddingSince every finite subgroup of the multiplicative complex numbers is cyclic,This is the cyclic-center obstruction to a faithful irreducible representation.

The complex regular representation has basis $(e_h{)}_{h\in G}$ and left actionIf $\rho (g)$ is the identity, thenso $g=1$. Thus $\mathbb{C}G$ is a faithful representation.

By Maschke's theorem, the regular representation decomposes as a direct sum of irreducibles,Its kernel is the intersection of the kernels of the constituent representations:Each constituent kernel is a normal subgroup of $G$. If $G$ is simple, every such kernel is either $\{1\}$ or $G$. They cannot all be $G$, because then their intersection would be $G$, contradicting faithfulness of the regular representation. Hence some $V_j$ has trivial kernel. Therefore every finite simple group has a faithful irreducible representation of a finite simple group.

Yes. If $g$ generates the cyclic group $C_n$, define the one-dimensional complex representationIt is irreducible because it is one-dimensional, and it is faithful because the displayed scalar has multiplicative order $n$. Thus every $C_n$, including the trivial case $n=1$, has a faithful irreducible representation.

Yes. WriteThe Faithful irreducible representation of D8 isThese matrices satisfy the defining relations. The two eigenspaces of $\rho (r)$ are exchanged by $\rho (s)$, so there is no common one-dimensional invariant subspace; the representation is irreducible. Its eight matrices $\rho (r{)}^j$ and $\rho (r{)}^j\rho (s)$ are distinct, so it is faithful.

No. The center of a group satisfiesbecause $Z(D_8)=\{1,r^2\}\cong C_2$. This centre is not cyclic. The cyclic-center obstruction to a faithful irreducible representation therefore rules out a faithful irreducible complex representation of $C_2\times D_8$.