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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.

Maschke's theorem says that every invariant subspace of a finite-dimensional complex representation of a finite group has an invariant complement. Average any projection over the group to make it equivariant. Induction on dimension therefore decomposes every representation into irreducibles.

The character of a representation is ${\chi }_V(g)=\mathrm{tr}(g{|}_V)$. For a $G$-set $X$, the permutation representation $\mathbb{C}X$ has basis $X$, and its character is the number of fixed points:For the regular action,The multiplicity of an irreducible $V_i$ is the character inner productso

If $V\cong {⨁}_i{n}_i{V}_i$, Schur lemma givesFor $V=\mathbb{C}G$, this is ${\sum }_i(\dim V_i{)}^2=|G|$.

Now suppose $\chi (g)=0$ off the identity. The multiplicity of $V_i$ in $V$ isThus $\dim V/|G|$ is a nonnegative integer and $V$ is that many copies of the regular representation. Finally,so