sum_g g^k, Frobenius-Schur indicators, S_n-invariants in freeAss(x_i), center of the group algebra

For any $k$ sums $T_k = 1/|G|\sum_{g\in G} g^k$ belong to the center of the group algebra, for finite group G. For $k=2$ they "are" (up to details and interpretation) Frobenius-Schur indicators. For $k>2$ similar things called "higher" FS indicators.

Question 1 Any element of the center of C[G], and $T_k$ in particular, acts by scalars in any irreducible representation. For $k=2$ the scalar can be $-1,0,1$ and brings important group theory information on the type of irrep (see WP and/or this MO question).

Is it true that values of higher $T_k$ in irreps are also integers ? Do they have some group theory interpretation ? (These MO questions 1, 2 discuss FS and relate them to dim of invariants on tensor powers of representation, but it seems different from my question.)

Question 1b (later edit)

One can generalize this consturction as follows consider free associative ring Z < x_1...x_|G| > . Consider S_|G| invariants in this algebra. Any such invariant can be mapped to the center of the group algebra, just substituting "g" instead of x_|g|. (see MO quest).
It will clearly give central element in C[G].

Is it true that there values will be interges in any irrep ? Moreover by analogy with T_k their values should be divisible by |G|.

Question 2

How far is the linear space/ algebra generated by $T_k$ from the whole center of group algebra? ( What can be said in general and in particular for S_n,A_n, GL_n(F_q), UT(n,q) )?

Same question not about T_k, but also about elements discussed in questions 1b above.

PS

Bonus Question 3 WP- article contains the following remark:

It resembles the Casimir invariants for Lie algebra irreducible representations.

In the sense that Casimirs are the center of U(g), while T_k are in the center of C[G].

Question: are there some more precise analogies ?

In some my research we studied the center of U(gl_n) and its loop algebra,
so such expression resembles "Gelfand" generators of U(gl_n) (which are traces of matrices "E^k", "E" given here), but I do not see any way to go beyond "resemble". While it would be nice seems now I am trying to think on some finite group questions.

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sciencedirect.com/science/article/pii/S0001870804000325 Central invariants and Frobenius–Schur indicators for semisimple quasi-Hopf algebras Geoffrey Masona, 1, , Siu-Hung Ng In this paper, we obtain a canonical central element νH for each semi-simple quasi-Hopf algebra H over any field k and prove that νH is invariant under gauge transformations. We show that if k is algebraically closed of characteristic zero then for any irreducible representation of H which affords the character takes only the values 0, 1 or −1, moreover if H is a Hopf algebra or a twisted quantum double of a –  Alexander Chervov Sep 16 '12 at 7:28

For question 1, they are integers, and this is related to Adams operations. The result basically follows from Newton's identities. However, there is no general bound on the size of the integers which may occur when $k >2.$
Later edit: In general, the $T_{k}$'s can fail very badly to generate the center of the group algebra. For example, if $p$ is an odd prime, and $G$ is a $p$-group of exponent $p,$ then $T_{k}$ is either $1_{G}$ (when $p$ divides $k$) or $\frac{1}{|G|} \sum_{g \in G}g$ (when $p$ does not divide $k$), whereas the dimension of the center of the group algebra of $G$ can be made as large as desired by choosing $G$ appropriately.
I just meant that when $k >2,$ there is no constant $c$ such that for every finite group $G$ and every irreducible character $\chi$ of $G,$ we have $\sum_{g \in G} \chi(g^{k}) \leq c|G|$. –  Geoff Robinson Sep 15 '12 at 17:10
Well, for example, when $k = 3$ (and similarly for any odd prime), whatever $c$ you choose, you can find a $3$-group $G$ and an irreducible character $\chi$ such that the sum of my previous comment exceeds $c|G|$. –  Geoff Robinson Sep 15 '12 at 18:54
Of course, doing this for $p=2$ shows there is no such $c$ for $k=4$, and then some direct products show there can be no $c$ for any value of $k>2$. –  Steve D Sep 15 '12 at 21:08