Squared finite-group irreducible dimensions sum to the group order

On the group basis of $\mathbb{C}[H]$, define $\lambda (h)\mathrm{\mid }x⟩=\mathrm{\mid }hx⟩$. Then $\lambda (h)\lambda (k)\mathrm{\mid }x⟩=\mathrm{\mid }hkx⟩=\lambda (hk)\mathrm{\mid }x⟩$ and $\lambda (e)=I$, so [F2] makes this a finite-dimensional representation. The regular representation is nonzero because $e\in H$. For any nonzero finite-dimensional representation, if it has no nonzero proper invariant subspace, [F3] says it is irreducible; otherwise choose such a subspace $W$ and [F1] supplies an invariant complement $U$. Both $W$ and $U$ are nonzero and have smaller dimension, so induction decomposes them into irreducibles. Applying this induction to $\mathbb{C}[H]$ proves that it is a finite direct sum of irreducible representations.

Let $\pi :H\to \mathrm{GL}(V)$ be any irreducible representation and choose $0\mathrm{\ne }v\in V$. Define the linear map $\mathrm{\Phi }:\mathbb{C}[H]\to V$ by $\mathrm{\Phi }({\sum }_h{a}_{h}h)={\sum }_h{a}_h\pi (h)v$. For $k\in H$, $\mathrm{\Phi }(k{\sum }_h{a}_{h}h)={\sum }_h{a}_h\pi (kh)v=\pi (k)\mathrm{\Phi }({\sum }_h{a}_{h}h)$, so $\mathrm{\Phi }$ is equivariant. Its image contains $v=\mathrm{\Phi }(e)$, is invariant, and therefore equals $V$ by irreducibility. By [F1], $\ker \mathrm{\Phi }$ has an invariant complement $W$ in $\mathbb{C}[H]$; then $\mathrm{\Phi }{\mathrm{\mid }}_W$ is onto and injective, so $W\cong V$ is an irreducible summand of $\mathbb{C}[H]$. At least one irreducible summand $U$ in the fixed decomposition of step 1.1 has $\mathrm{\Phi }{\mathrm{\mid }}_U\mathrm{\ne }0$, since $\mathrm{\Phi }$ is onto a nonzero space. Its kernel and image are invariant, so irreducibility of $U$ and $V$ makes $\mathrm{\Phi }{\mathrm{\mid }}_U$ an isomorphism. Consequently every irreducible representation occurs among those finitely many summands.

For each irreducible representation $\pi$, choose any positive-definite inner product and average it as $⟨v,w{⟩}_H:={\sum }_{h\in H}⟨\pi (h)v,\pi (h)w⟩$. This is positive definite and is $H$-invariant by reindexing the finite sum, so $\pi (h)$ is unitary and ${\chi }_{\pi }(h^{-1})=\overline{{\chi }_{\pi }(h)}$. For two irreducibles $\pi ,\sigma$, define $P(T):=\mathrm{\mid }H{\mathrm{\mid }}^{-1}{\sum }_{h\in H}\sigma (h)T\pi (h^{-1})$ on ${\mathrm{Hom}}_{\mathbb{C}}(V_{\pi },V_{\sigma })$. Reindexing gives $\sigma (k)P(T)\pi (k^{-1})=P(T)$ for every $k$, so $P(T)$ intertwines the group actions; if $T$ already intertwines, every summand equals $T$. Hence $P$ is a projection with image exactly ${\mathrm{Hom}}_H(V_{\pi },V_{\sigma })$.

For matrices $A=\sigma (h)$ and $B=\pi (h^{-1})$ in chosen bases, the coefficient of a rectangular matrix unit $E_{ij}$ in $A{E}_{ij}B$ is $A_{ii}{B}_{jj}$. Summing these diagonal coefficients gives $\mathrm{tr}(T\mapsto ATB)={\sum }_{i,j}{A}_{ii}{B}_{jj}=\mathrm{tr}A\mathrm{tr}B={\chi }_{\sigma }(h)\overline{{\chi }_{\pi }(h)}$. Linearity of the diagonal sum yields $\mathrm{tr}P=\mathrm{\mid }H{\mathrm{\mid }}^{-1}{\sum }_h{\chi }_{\sigma }(h)\overline{{\chi }_{\pi }(h)}$. Since $P^2=P$, every $T$ splits as $P(T)+(T-P(T))\in \mathrm{im}P\oplus \ker P$; the intersection is zero because $P$ is the identity on its image. In a basis adapted to this direct sum, $P$ has matrix $\mathrm{diag}(I,0)$, so $\mathrm{tr}P=\dim \mathrm{im}P$. An intertwiner between irreducibles is either zero or an isomorphism because its kernel and image are invariant. By [F4]–[F5], $\mathbb{C}$ is algebraically closed, so [F6] gives every endomorphism $T$ of the nonzero finite-dimensional space $V_{\pi }$ an eigenvalue $\alpha$. For $T\in {\mathrm{End}}_H(V_{\pi })$, the eigenspace $\ker (T-\alpha I)$ is nonzero and $H$-invariant because $T$ intertwines the $H$-action; irreducibility forces it to be all of $V_{\pi }$, so $T=\alpha I$. Thus $\dim \mathrm{im}P$ is $1$ if $\pi \cong \sigma$ and $0$ otherwise, proving character orthogonality by the averaging calculation.

The character of $\lambda$ is ${\chi }_{\lambda }(e)=\mathrm{\mid }H\mathrm{\mid }$ and ${\chi }_{\lambda }(h)=0$ for $h\mathrm{\ne }e$, since left multiplication by $h\mathrm{\ne }e$ fixes no group-basis element. Define $⟨\alpha ,\beta ⟩:=\mathrm{\mid }H{\mathrm{\mid }}^{-1}{\sum }_h\overline{\alpha (h)}\beta (h)$. If $\lambda \cong {⨁}_{\pi }{m}_{\pi }\pi$ as in step 1.1, then character additivity and the orthogonality from step 3.1 give $m_{\pi }=⟨{\chi }_{\pi },{\chi }_{\lambda }⟩=\mathrm{\mid }H{\mathrm{\mid }}^{-1}{\sum }_h\overline{{\chi }_{\pi }(h)}{\chi }_{\lambda }(h)=\dim \pi$.

Taking dimensions of the regular decomposition yields $\mathrm{\mid }H\mathrm{\mid }=\dim \mathbb{C}[H]={\sum }_{\pi }{m}_{\pi }\dim \pi ={\sum }_{\pi }(\dim \pi {)}^2$. This proves the stated formula and finiteness.