Sum of squares of degrees

For a finite group, the sum of the squares of the dimensions of its irreducible complex representations equals the group order.

Sum of squares of degrees

Let $G$ be a finite group and let $k$ be an algebraically closed field with $\operatorname{char}(k)\nmid |G|$ (in particular, $k=\mathbb C$ ). Let

\{V_1,\dots,V_r\}

be a complete set of pairwise non-isomorphic finite-dimensional irreducible representations of $G$ over $k$ , and write $d_i=\dim_k(V_i)$ .

Theorem (sum of squares of degrees)

\sum_{i=1}^r d_i^2 \;=\; |G|.

Equivalently, the underlying $kG$ -module of the regular representation decomposes as

k[G]\;\cong\;\bigoplus_{i=1}^r d_i\,V_i,

and taking dimensions gives $\dim_k(k[G])=|G|=\sum_i d_i\dim(V_i)=\sum_i d_i^2$ .

This decomposition is a standard consequence of Maschke-type semisimplicity statements (so $k[G]$ is semisimple) together with multiplicity computations using Schur's lemma .

Cyclic group $C_n$ .
Over $\mathbb C$ , every irreducible representation of $C_n$ is $1$ -dimensional (a character). There are $n$ such characters, so
$\sum_i d_i^2 \;=\; \underbrace{1^2+\cdots+1^2}_{n\text{ times}} \;=\; n \;=\; |C_n|.$
Symmetric group $S_3$ (order $6$ ).
$S_3$ has irreducible degrees $1,1,2$ (trivial, sign, and the $2$ -dimensional standard representation), hence
$1^2+1^2+2^2 \;=\; 6 \;=\; |S_3|.$
Dihedral group $D_8$ (symmetries of a square, order $8$ ).
$D_8$ has four $1$ -dimensional irreducibles and one $2$ -dimensional irreducible, so
$4\cdot 1^2 + 1\cdot 2^2 \;=\; 8 \;=\; |D_8|.$