# Changes

## Sum of squares of degrees of irreducible representations equals order of group

, 03:59, 13 July 2011
Proof in characteristic zero
| 3 || $\alpha$ is the sum $\sum_{i=1}^r d_i\chi_i$ || Facts (1),(2) || $\chi_i$ are characters of (all the) irreducible representations. || Step (2) || <toggledisplay>We know by Fact (1) that <matH>\alpha[/itex] is a nonnegative integer combination of the $\chi_i$s, because $\rho$ is a nonnegative integer combination of the corresponding representations. By Fact (2) and Step (2), the coefficients are precisely $\langle \alpha, \chi_i \rangle = d_i$.</toggledisplay>
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| 4 || The value of $\alpha$ iat at the identity element is $\sum_{i=1}^r d_i^2$. || || || Step (3) || <toggledisplay>At the identity element, $\chi_i$ takes the value $d_i$, so plugging in Step (3) gives this.</toggledisplay>
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| 5 || $\sum_{i=1}^r d_i^2 = |G|$ || || || Steps (1), (4) || <toggledisplay>By Step (1), $\alpha$ at the identity element of $G$ is $|G|$. By Step (4), it is $\sum_{i=1}^r d_i^2$. Combining, we get the result.</toggledisplay>
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===Proof in other characteristics===
This follows from the characteristic zero proof, and the fact that [[degrees of irreducible representations are the same for all splitting fields]].