# Does the monoid of non-zero representations with the tensor product admit unique factorization?

Let $$(M, \cdot, 1)$$ be a monoid. We will now define the notion of unique factorization monoid. A non-invertible element in $$M$$ is called irreducible if it cannot be written as the product of two other non-invertible elements. $$M$$ is called a unique factorization monoid if every non-invertible element $$m$$ admits a factorization $$m = n_1 \cdot \ldots \cdot n_k$$ into irreducibles which is unique up to reordering and multiplication by invertible elements.

Let $$G$$ be a finite group. Let $$\operatorname{Rep}$$ be a set containing precisely one representative from each isomorphism class of finite-dimensional complex representations of $$G$$.

$$(\operatorname{Rep}, \oplus, \{0\})$$ is a monoid and the fact that this monoid is a unique factorization monoid is precisely the fact that every finite-dimensional complex representation of $$G$$ splits uniquely into a direct sum of irreducibles.

Let $$\operatorname{Rep}^* = \operatorname{Rep} \setminus \{\{0\}\}$$. Now, $$(\operatorname{Rep}^*, \otimes, \mathbb{C})$$ is a monoid. Is this monoid necessarily a unique factorization monoid?

Counterexamples are much easier to produce than this and exist already when $$G = C_2$$ and with two $$2$$-dimensional representations, see here.

Abstractly the problem is that the representation ring is not a domain, so there's no reason one should be able to cancel factors from a tensor product. In general the representation ring of $$G$$ over $$\mathbb{C}$$ is isomorphic to the algebra of class functions $$G/G \to \mathbb{C}$$ so as a ring it is the product $$\mathbb{C}^{c(G)}$$ where $$c(G)$$ is the number of conjugacy classes, so it has many zero divisors.

• What do you mean by "$G/G \to \mathbb{C}$"? Also, doesn't this confuse the representation ring $R(G)$ with the algebra $R(G) \otimes \mathbb{C}$? Commented Jun 22 at 20:33
• @Smiley1000: here $G/G$ means the quotient of $G$ by the action of $G$ by conjugation, so it denotes the set of conjugacy classes. Then a function $G/G \to \mathbb{C}$ is a complex-valued function on the set of conjugacy classes. By "representation ring over $\mathbb{C}$" I meant $R(G) \otimes \mathbb{C}$, I guess that was ambiguous. Commented Jun 22 at 20:35
• Thanks. Although I must admit that writing $G/G$ and implying the action by conjugation is something I've never seen before, perhaps also a little too ambiguous. Commented Jun 22 at 20:37
• It's notation that gets used in some corners of geometric representation theory. There people do things like take $G$ to be an algebraic group and consider the quotient $G/G$ in a stacky sense. Commented Jun 22 at 20:38

A counterexample is given by Nate at https://math.stackexchange.com/a/4436073/491450 :

Taking $$G = A_5$$, we have $$V_4 \otimes V_5 \otimes V_3 \cong V_4 \otimes V_5 \otimes {V_3}'$$ where the subscript denotes the dimension of the representation. However, $$V_3$$ and $${V_3}'$$ cannot be transformed into one another by tensoring with some one-dimensional representation (the only one-dimensional representation is trivial).