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Given an (irreducible complex) character of a finite group the following three construction all yield another irreducible character of the same degree:

  1. multiplying by a degree 1 character
  2. applying an outer automorphism
  3. taking a Galois conjugate

Note that 2) and 3) never change the set of character values, they just permute the list.

The degree five characters of $ A_6 $ are not Galois conjugate but are related by an outer automorphism. The two degree $ 16 $ characters of $ M_{11} $ are not related by an outer automorphism but are Galois conjugate.

What is an example of a finite group $ G $ and two distinct irreducible characters of $ G $ which have the same character values but are not related by any combination of the three constructions given above?

This is cross posted from MSE where it has been up for about a week with 6 upvotes but no comments or answers:

https://math.stackexchange.com/questions/4856291/distinct-characters-with-the-same-character-values-outer-automorphisms-and-galo

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1 Answer 1

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Take $G = S_3 \times S_4$ and consider the unique two-dimensional irreducible representation of $S_3$ and the unique two-dimensional irreducible representation of $S_4$. These have the same character values $2,0,-1$ (in fact they are both pullbacks of the same representation under two different homomorphisms $S_4 \to S_3$ ) but are not related by any combination of the operations.

Galois conjugation is irrelevant since the characters are rational. Twisting by one-dimensional characters can only twist by quadratic characters and doesn't affect the value on order $3$ elements - i.e. $3$-cycles in $S_3$, $3$-cycles in $S_4$, or products of $3$-cycles in both $S_3$ and $S_4$. So it remains to consider outer automorphisms, which would have to permute these three conjugacy classes. But no outer automorphism can since their centralizers have different orders: $3 \cdot 24, 6 \cdot 3,$ and $3 \cdot 3$ respectively.

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  • $\begingroup$ would you be willing to cross post this to the original MSE question so I can accept it there as well? $\endgroup$ Commented Feb 9 at 19:48

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