Existence of irrep containing a particular element for a finite group

Question

Let $ G $ be a finite group.

Idea: Can any matrix that "looks like" it belongs in $ G $ be found in the image of some faithful irrep of $ G $?

Formal statement of question:

Let $ G $ be a finite group. Let $ \chi $ be a faithful irreducible degree $ d $ character of $ G $.

Let $ V $ be a $ d \times d $ unitary matrix of order $ n $.

If $ G $ has a conjugacy class of elements of order $ n $, call this conjugacy class $ [g] $, such that $ \chi([g])=Trace(V) $ then does there always exist a faithful irrep $$ \pi: G \to \mathrm{U}(n) $$ with character $ \chi $ and such that $ V $ is in the image of $ \pi $?

Thoughts: My guess is that this isn't quite true as stated, but is mostly true. Maybe adding more constraints on the relationship of $ V $ to $ \chi $ and $ G $ will help.

Edit: specified that representations should be irreducible, I don't think that will change the answer to the question

Answer 1

I'm ignoring unitary -- for finite groups it always can be enforced, but for me it just makes examples more complicated.

If we look at single elements in the group, we can alway assume a basis in which the element is diagonal, but Eigenvalues of course come from the irreducible constiutents.

Thus, if the eigenvalues of $V$, with multiplicities, can be combined from multiplicities of the same eigenvalues on the chosen class for certain collections of irreducible representations, then we can force $V$ in $\pi(G)$: Simply takethe direct sum of these representations, all in basis so they are diagonal on the particular class.

Otherwise not. Counterexamle $A_5$, order 3. $\zeta=e^{(2\pi i)/3}$.

Let's find the multiplicities of the eigenvalues for the different representations. In GAP:

gap> g:=AlternatingGroup(5);;
gap> List(ConjugacyClasses(g),x->Order(Representative(x)));
[ 1, 2, 3, 5, 5 ]
gap> pos:=3;
3
gap> List(Irr(g),x->EigenvaluesChar(CharacterTable(g),x,pos));
[ [ 0, 0, 1 ], [ 1, 1, 1 ], [ 1, 1, 1 ], [ 1, 1, 2 ], [ 2, 2, 1 ] ]

We see that $\zeta$ and $\zeta^2$ always ocurr with the same multiplicity. (Actually, we could have seen this from the fact that the column for order 3 in the character table is rational.)

Thus a scalar matrix $V=\zeta\cdot I_d$ (in whatever dimension $d$), obviously of order $3$, can never be in the image of any representation of $A_5$.