In the book by Fulton and Harris they make the following claim: All one-dimensional representations of a finite group $G$ are trivial on proper normal subgroups whose quotient group is cyclic. My only idea is the following: Let $(\rho,V)$ be a representation of $G$. Then, $\rho$ is trivial on $H$ (a normal subgroup) if and only if $\rho$ factors as $\bar{\rho}$ through the quotient $G/H$. Since $G/H$ is cyclic it is enough to define $\bar{\rho}$ in the generator $gH$. Let $\bar{\rho}(gH) = \rho(g)$. However, I'm not able to prove that this is well defined. I know that somehow I have to use the fact that $\rho$ is one-dimensional, that is, its image is abelian.

PS: This is used to prove that $S_5$ has only two non-trivial one-dimensional representations. However, in this specific case $A_5$ is not only a 'normal subgroup whose quotient is cyclic', it is the derived subgroup of $S_5$. Therefore, every group homomorphism with abelian image is trivial on $A_5$. The claim follows from the fact that a one-dimensional representation has abelian image. I don't know where the 'cyclic' hypothesis part enters.


1 Answer 1


The "claim" is false.

Counterexample. Consider the group $G = H = \langle\, a \mid a^2 \,\rangle$ of order two. Obviously, $G/H = 1$ is cyclic. As a one-dimensional representation (over $\mathbb{C}$), take $$ \rho \colon G \to \mathbb{C}^\ast;\quad a \mapsto -1. $$

  • $\begingroup$ do you have a proper counterexample? I edited the statement for proper normal subgroups $\endgroup$
    – user512346
    Commented Apr 4, 2019 at 11:12
  • $\begingroup$ Just take a faithful representation of a cyclic group of order 4. This is nontrivial on the subgroup of order 2. $\endgroup$
    – Derek Holt
    Commented Apr 4, 2019 at 11:22
  • $\begingroup$ Shoot. That was a stupid question. Thanks. $\endgroup$
    – user512346
    Commented Apr 4, 2019 at 11:31

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