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The sum of the degrees of the irreducible complex characters (not the square sum which is the group order) is relevant do determine the dimension of a maximal torus in the group algebra.

I have investigated certain group classes as extra-special groups, abelian groups, direct products, quaternion groups, dihidral groups, semidih, groups, $S_4$ and $A_4$, $SL(2,q)$, $GL(2,q)$, meta-zyclic groups, met-abelian-groups, central products, frobenius groups and determine this sum. Also symmetric groups are no known (thanks to the answer of Alex).

My questions is whether there is a nice sum-formula (maybe with recursion) for groups like: alternating groups, p-groups (maybe special classes), simple groups, general linear, special linear groups, groups with united factor group, groups with normal p-Sylow-subgroup and abelian Hall-Complement ... as well.

In the literature there is a nice inequation saying that this sum is at least the double of the linear characters for a non-soluble groups (using the classification atlas). Maybe there are some more upper and lower bounds to derive here.

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For those groups whose complex representations can all be defined over the reals, such as the symmetric groups and the dihedral groups for example, the sum of the degrees of the characters is exactly equal to 1 + the number of elements of order 2. This follows from the theory of Frobenius-Schur indicators. See here and here for example.

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  • $\begingroup$ Thanks for your hint on counting involutions for groups having only real valued irreducible characters. Do you know if there is a classification of such groups or do you know a list of examples besides symmetric and dihedral groups? $\endgroup$
    – Sven Wirsing
    Sep 2, 2014 at 21:21
  • $\begingroup$ I have asked the whole topic on mathoverflow as well. $\endgroup$
    – Sven Wirsing
    Sep 4, 2014 at 7:40
  • $\begingroup$ There are some news about this question on mathoverflow: mathoverflow.net/questions/180046/… $\endgroup$
    – Sven Wirsing
    Sep 4, 2014 at 18:42
  • $\begingroup$ I flag this as answered combined with the answered in mathoverflow! $\endgroup$
    – Sven Wirsing
    Sep 6, 2014 at 7:12

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