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Let $G$ be a finite group and $\pi$ an irreducible complex representation. The Frobenius-Schur indicator of $\pi$ is defined as:
$$ \nu_2(\pi):=\frac{1}{|G|} \sum_{g \in G} \chi_{\pi}(g^2) $$ with $\chi_{\pi}$ the character of $\pi$.

Note that the map $s: g \mapsto g^2$ is well-defined on the conjugacy classes as $\tilde{s}: C(g) \mapsto C(g^2)$ because $(hgh^{-1})^2 = hg^2h^{-1}$. Let $\chi_1, \cdots, \chi_r$ be the irreducible characters of $G$ (with $\chi_i = \chi_{\pi_i}$), and $C_1, \cdots, C_r$ be the conjugacy classes, with $\chi_1$ the trivial and $C_1 = C(1)$. The character table of $G$ is given by the matrix $(\chi_{i,j})$ with $\{ \chi_{i,j} \} = \chi_i(C_j)$. The map $\tilde{s}$ induces a map $m$ on $\{1,2, \cdots, r \}$ such that $\tilde{s}(C_j) = C_{m(j)}$. It follows that the Frobenius-Schur indicator $\nu_2$ is completely determined by the character table $(\chi_{i,j})$ and the map $m$ as follows:

$$ \nu_2(\pi_i):=\frac{1}{|G|} \sum_{j} |C_{j}|\chi_{i,m(j)} = \sum_j \frac{\chi_{i,m(j)}}{\sum_i |\chi_{i,j}|^2}$$ because $|C_j| = |G|/\sum_i |\chi_{i,j}|^2$.

Note that the character table alone is not sufficient to determine $\nu_2$. For example, the quaternion group $Q_8$ and the dihedral group $D_4$ have same character table, but the first admits an irreducible complex representation with Frobenius-Schur indicator $-1$ (in fact it is the smallest such finite group) whereas the second not. But these do not have the same class type $(1,2,4A,4B,4)$ for the first and $(1,2A,2B,2C,4)$ for the second (a class is of type $nX$ if its elements has order $n$).

Question: Is the Frobenius-Schur indicator $\nu_2$ completely determined by the character table including the class types? If so, what is the formula?

It is "suggested" true by the section 71.12-5 in GAP manual, as GAP seems to need these data only to compute $\nu_2$.

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    $\begingroup$ I would guess the answer is no, I guess you need to know the power map $g \mapsto g^2$ to compute $\nu_2(\chi)$. About GAP, aren't the power maps usually included in the character table? With Display(CharacterTable("A5"));; you see the character table of $A_5$ along with the power maps. Perhaps with a computer search you could find two groups with same character tables and class types, but different Frobenius-Schur indicators. $\endgroup$ – spin Sep 19 at 13:01
  • $\begingroup$ @spin: two non-isomorphic groups with same character table and class types are called a Brauer pair. Among the $2$-groups, the smallest order of a group in a Brauer pair is $2^8$, and among the $56092$ groups of order $2^8$, there are exactly ten Brauer pairs (see MR2680716 Theorem 2.6.2 page 136). Perhaps the counter-example you expect is among these ten pairs. $\endgroup$ – Sebastien Palcoux Sep 19 at 14:46
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    $\begingroup$ Is that the usual definition of a Brauer pair? I thought Brauer pairs also have the same power map, hence they would have the same FS-indicators as well. $\endgroup$ – spin Sep 19 at 17:44
  • $\begingroup$ @spin: You are right, thanks! $\endgroup$ – Sebastien Palcoux Sep 19 at 20:38
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I don't see a way at the moment to a full answer, but I mention the following in case someone can make use of it : the class function Sqr defined by Sqr(g) = the number of square roots of G in G is always a generalzied character: we have ${\rm Sqr}(g) = \sum_{ \chi \in {\rm Irr}(G)} \nu_{2}(\chi) \chi(g)$ for each $g \in G$, and clearly Sqr contains the trivial character with multiplicity one.

It follows that a necessary and sufficient condition for there to be no irreducible character $\chi$ of $G$ with $\nu_{2}(\chi) = -1$ is that the function Sqr is a genuine character of $G$, that is, a non-negative integer (not all zero) combination of irreducible characters of $G$.

(continued..) In general, I don't think it's clear how to calculate the function Sqr just from the class types and character table: in the examples given of the quaternion and dihedral groups of order 8 it is easy, because elements of order $4$ have no square root (in these groups), while in $Q8$ the identity has only two square roots and an element of order $2$ has six square roots , and in $D8$ the central involution has two square roots, the identity has $6$ square roots, and the non-central involutions have no square roots.

In general, when there are many classes of elements of the same order, I think it is less clear how to calculate just from class types and character table how many square roots elements have.

Conversely, in groups $G$ where Sqr is determinable from this information, we can tell whether Sqr is a character by calculating $\langle {\rm Sqr}, \chi \rangle $ for each irreducible character $\chi$.

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From the source code of GAP:

#############################################################################
##
#M  IndicatorOp( <ordtbl>, <characters>, <n> )
#M  IndicatorOp( <modtbl>, <characters>, 2 )
##
InstallMethod( IndicatorOp,
    "for an ord. character table, a hom. list, and a pos. integer",
    [ IsOrdinaryTable, IsHomogeneousList, IsPosInt ],
    function( tbl, characters, n )
    local principal, map;

    principal:= List( [ 1 .. NrConjugacyClasses( tbl ) ], x -> 1 );
    map:= PowerMap( tbl, n );
    return List( characters,
                 chi -> ScalarProduct( tbl, chi{ map }, principal ) );
    end );

So in the computation performed on a character table, GAP computes the power map $g \mapsto g^n$ on the conjugacy classes of $G$, and with this for each character $\chi$ the scalar product $\nu_n(\chi) = \langle \chi^{(n)}, 1 \rangle$, where $\chi^{(n)}(g) = \chi(g^n)$ for all $g \in G$.

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  • $\begingroup$ This shows that the suggested informal justification doesn't justify, rather than answering the main question ("Is $\nu_2$ determined by …") yes or no, right? $\endgroup$ – LSpice Sep 19 at 14:52
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    $\begingroup$ @LSpice: Yes, to be more clear my answer doesn't settle the main question posed by Sebastien. Point is that GAP does use more information than just the character table and class types to compute Frobenius-Schur indicators. $\endgroup$ – spin Sep 19 at 16:57

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