Let $G$ be a finite group. A basic result in representation theory is that up to $\mathbb{C}[G]$-module isomorphism, there are only finitely many irreducible representations of $G$ over $\mathbb{C}$. The way I'm familiar with proving this is to let $(\pi, W)$ be any irreducible representation of $G$ with character $\chi$, and let $(\phi, V)$ be the left regular representation of $G$ ($V$ is the vector space with formal basis $x_g : g \in G$, and the $\mathbb{C}[G]$-module structure is given by $gx_h = x_{gh}$) with character $\gamma$. One can prove that "the number of times $W$ occurs in $V$" is equal to $$(\chi, \gamma) = \frac{1}{|G|} \sum\limits_{g \in G} \chi(g) \overline{\gamma(g)}$$ and one can then quickly argue that this is not zero.

It follows that every irreducible representation of $G$ occurs as a direct summand in $V$, and from the limited uniqueness a representation has as a direct sum of irreducible subrepresentations, you can see that there are only finitely many irreducible representations up to isomorphism

However, is there a more intuitive way of seeing that there are only finitely many irreducible representations of $G$? Without any character theory, I know that if $(\pi,W)$ is irreducible, then the dimension of $W$ must be $\leq$ the order of $G$, because if $0 \neq v \in W$, then the span of $gv : g \in G$ must be all of $W$. I am having trouble coming up with any immediate results beyond that.

  • $\begingroup$ The first way I learned how to prove this is the orthogonality relations: once you know that the characters of irreps are orthogonal, it immeiately follows that there are at most as many irreps as the dimension of the space of class functions. But it sounds like you're trying to avoid character theory. $\endgroup$ Dec 22, 2015 at 6:23

2 Answers 2


Let $\rho$ be a representation of $G$ on the vector space $V$. For any $v \in V$ and $l$ linear functional on $V$ consider the "matrix coefficient" $\phi_{l,v}$, a function from $G$ to $k$ defined as follows $$\phi_{l,v}(g) = l (g v)$$

The equality $$\phi_{l,hv}(g) = l (g hv)= \phi_{l,v}(gh)$$ implies that the map from $V$ to $k$ valued functions on $G$, $$v\mapsto \phi_{l,v}(\cdot)$$ is a morphism of representations. Assume now that $V$ is irreducible and $l\ne 0$. Then we conclude that $V$ imbeds into the space of $k$ valued functions on $G$, that you may call $k[G]$. Thus, we see that every irreducible representation is a subrepresentation of $k[G]$.

Assume now that $G$ is finite. Let us show that the sum of the dimensions of irreducible representations is $\le |G|$. Let $V_1$, $\ldots$, $V_N$ irreducible subrepresentations of $k[G]$ so that $\sum \dim V_i > |G|$. Let $i$ minimum so that the sum of $V_1$, $\ldots$, $V_i$ is not direct. Then $V_i \subset V_1 \oplus \cdots \oplus V_{i-1}$ and so it will be isomorphic to one of the $V_1$, $\ldots$, $V_{i-1}$.

  • $\begingroup$ Took me a second to see why $v \mapsto \phi_{l, v}$ is injective: if $\phi_{l, v} = 0$ for some $v \neq 0$ then $l(gv) = 0$ for all $g \in G$, i.e. $l(Gv) = 0$. Since $V$ is irreducible, $Gv = V$, so in other words $l(V) = 0$, but this contradicts the assumption that $l \neq 0$. $\endgroup$ Dec 22, 2015 at 7:27
  • $\begingroup$ @DanielMcLaury: Indeed! It's also good to notice that the translates of any non-zero matrix coefficient span an irreducible representation isomorphic to $V$ inside $k[G]$. $\endgroup$
    – orangeskid
    Dec 22, 2015 at 7:38

The proof of the Artin-Wedderburn theorem (together with Maschke's theorem) shows that $\mathbb{C}[G]$ is a finite direct product of matrix algebras $M_{d_i}(\mathbb{C})$, one for each irreducible representation of dimension $d_i$. In particular, it follows immediately, with no character theory, that the number of irreducible representations is at most $|G|$ (and even, again with no character theory, that $|G| = \sum_i d_i^2$).

More generally, if $A$ is a finite-dimensional algebra over $\mathbb{C}$, simple modules of $A$ are the same as simple modules of the quotient $A/J(A)$ where $J(A)$ denotes the Jacobson radical. This ring is semisimple, and applying Artin-Wedderburn to it as above again implies that the number of simple modules is at most $\dim A$.

You can consider these results as noncommutative analogues of the claim that a polynomial has finitely many roots, which you get by applying the above argument to commutative algebras of the form $\mathbb{C}[x]/f(x)$. The two results even overlap in the case of cyclic groups, which you get by setting $f(x) = x^n - 1$.

Alternatively, if you're willing to believe that a representation is determined by its character, and that an irrep has dimension at most $|G|$, you can argue (without orthogonality) that there are finitely many irreducible characters by bounding the number of possible values an irreducible character in each dimension can have: namely, if $V$ is an $n$-dimensional irrep, then the character $\chi_V(g)$ is a sum of $\dim V$ roots of unity of order dividing the order of $g$.


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