# Which finite groups have characters injective on conjugacy classes?

This is an idle question inspired by wondering if one could further generalise @ClarkLyons's generalisation of an answer of @diracdeltafunk.

Of course, every finite group has a faithful, finite-dimensional representation (say, its regular representation), which is to say, every finite group admits an injective homomorphism into some $$\operatorname{GL}_n(\mathbb C)$$. (I'm used to complex representations, but we can consider the regular representation over any field.)

For which finite groups $$G$$ is there a representation $$\pi : G \to \operatorname{GL}_n(\mathbb C)$$ such that $$g \mapsto \operatorname{tr} \pi(g)$$ is injective on conjugacy classes? How much does the answer change if we replace $$\mathbb C$$ by the algebraic closure of $$\mathbb F_p$$?

Proposition: If $$G$$ is a finite group then there is a homomorphism $$\pi:G \to \operatorname{GL}_n(\mathbb{C})$$ for some positive integer $$n$$ so that $$\chi:G \to \mathbb{C}:g \mapsto \operatorname{tr}(\pi(g))$$ is injective on conjugacy classes of $$G$$.

I'll use the following two facts from character theory (chapter 2 and 3 of Isaacs's textbook on the Character Theory of Finite Groups):

Lemma: Two elements $$g,h$$ of $$G$$ are conjugate in $$G$$ if and only if $$\chi(g) = \chi(h)$$ for all irreducible characters $$\chi$$ of $$G$$

Lemma: $$|\chi(g)| \leq \chi(1)$$, $$\chi(g)$$ is an algebraic integer in $$\mathbb{Z}[\zeta_m]$$, and $$\chi(1)$$ is an integer dividing $$m=|G|$$.

Now the proof converts irreducible characters to the $$N$$-adic digits of a larger character.

Proof: Let $$N > |G|$$ and define $$\chi = \theta_1 = \sum_{i=1}^k N^{i-1} \chi_i$$ to be a linear combination of the $$k$$ irreducible characters of $$G$$.

If $$\chi(g) = \chi(h)$$ then the same is true mod $$N$$ in the ring of algebraic integers $$\mathbb{Z}[\zeta_m]$$, but then $$\chi_1(g) = \chi_1(h)$$ since the rest of $$\chi$$ is a multiple of $$N$$.

Now consider the remainder, $$\theta_2=(\chi-\chi_1)/N = \sum_{i=2}^k N^{i-2} \chi_i$$. Since both $$\chi(g) = \chi(h)$$ and $$\chi_1(g)=\chi_1(h)$$, we have $$\theta_2(g)=\theta_2(h)$$. Continue as before until we get $$\chi_k(g)=\chi_k(h)$$ and so by the lemma $$g,h$$ are conjugate in $$G$$. ∎

• Let me know if you want any of the character theory translated. Most of it is just "the eigenvalues of $\pi(g)$ are $n$th roots of unity where $n=|G|$" May 29, 2021 at 1:17
• Correct, I'll fix that May 29, 2021 at 1:33
• I suppose that the same trick will allow us to handle mod-$p$ representations of groups of order prime to $p$; and then the characters of such representations cannot distinguish an element of order $p$ from the identity, so the mod-$p$ question is also answered. May 29, 2021 at 1:44
• Yes. Usually for mod p instead of the trace we use the Brauer character and p-regular classes. The p-regular classes takes care of the eigenvalues being identical by not talking about elements whose order is divisible by p. The Brauer character takes care of distinguishing when sums of eigenvalues are equivalent mod p. For example, if the eigenvalues are 1,1,1,1 versus 1,2,2,2 (all mod p=3), then the trace is 1+1+1+1=1 versus 1+2+2+2=1 (all mod p=3). The Brauer character is 1+1+1+1=4 versus 1-1-1-1=-2, both as algebraic integers, not mod anything. May 29, 2021 at 12:31