This is not homework.

Motivation : I have been reading "Representation and characters of groups" by James & Liebeck, and in chapter 9 they introduce Schur's Lemma, which states that for a finite group $G$, any $\mathbb CG$-homomorphism between two irreducible finite dimensional $\mathbb C G$-modules is of the form $\lambda I_n$ with $I_n$ the identity matrix of order $n$ ($\lambda$ is possibly $0$).

Right after it, they state that every irreducible representation of an abelian group $G$ to $GL_n(\mathbb C)$ is such that $n=1$, because every vector subspace of the $\mathbb C G$-module will be a submodule in the abelian case (implicitly using Maschke's theorem here to complete the proof).

Now is the part that interested me ; using the fact that $G$ is isomorphic to a finite direct product of finite cyclic groups, one can deduce that any representation of $G = \langle g_1, \dots, g_n \rangle$ will have the form $\rho : G \to \mathbb C$ with $\rho(g_1^{i_1} \dots g_n^{i_n}) = \lambda_1^{i_1} \dots \lambda_n^{i_n}$, with the $\lambda$'s being roots of unity of the order of their respective generator. There are $|G|$ such representations (depending of the choice of the $\lambda$'s). I don't want to enter in the details of this.

Question : Can we build a converse for this theorem? Namely, I'm not sure what would be all the prerequisites, but say that $G$ is finite and that every representation of $G$ has dimension $1$ and is "root of unity-valued", can one deduce that $G$ is abelian and isomorphic to a direct product of cyclic groups using representation theory?

What I've attempted to say : Up to now I always needed the fact that $G$ was abelian ; my attempt was to get a faithful irreducible representation of $G$, and then produce an isomorphism between $G$ and the induced group of roots of unity that this representation produced. Then I would work over $\mathbb C$ and (I assume that it is possible that) I am done.

For instance ; there always exists one representation of $G$ that is faithful, namely the permutation module of $G$ over $\mathbb C^{|G|}$. But I don't know if I can get an irreducible representation from there ; I know that this representation will always be reducible if $|G| > 1$, but when reducing using Maschke's theorem I feel like I'm going to lose the faithful property. Can I just assume that there exists a faithful irreducible representation of $G$ for some nice reason?

Any clarification on some steps of the proof will be appreciated as an answer. Thanks

  • $\begingroup$ Take a look at this: math.uconn.edu/~kconrad/blurbs/grouptheory/charthy.pdf $\endgroup$ Mar 10, 2012 at 1:26
  • $\begingroup$ If $V$ is a faithful representation of a finite group $G$, and you are working over the complex numbers say, where you know Maschke's theorem holds, then it follows that $V=U_1\oplus U_2\oplus \ldots\oplus U_k$ for some irreducible subrepresentations $U_1,\ldots,U_k$ of $V$. You are right that the individual subreps $U_i$ may not be faithful, but this does not mean you lose anything: the kernel $K_V$ of the action of $G$ on $V$ is equal to $\bigcap_{i=1}^k K_i$ where $K_i$ is the kernel of the action of $G$ on $U_i$. $\endgroup$
    – krm2233
    Jul 18 at 12:37

1 Answer 1


The converse is true because diagonal matrices commute:

If you look at the $G$-module $\mathbb{C}[G]$, Maschke's theorem gives you a basis $\mathcal{B}$ for $\mathbb{C}[G]$ in which

$$\text{mat}_{\mathcal{B}}(\pi(g)) = \left(\begin{matrix}\chi_1(g) & &\\& \ddots&\\&&\chi_n(g)\end{matrix}\right)$$

Where $\chi_1,\ldots, \chi_n$ are characters of $G$ (where we denote $\pi(g) : \mathbb{C}[G] \to \mathbb{C}[G]$ the action of $g \in G$ on that space). Now, because the module is faithful, the morphism $g \mapsto \text{mat}_{\mathcal{B}}(\pi(g))$ from $G$ to $\text{GL}_n(\mathbb{C})$ is injective. But obviously the image of this map is a commutative group, so $G$ is also commutative.


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