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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?

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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.

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  • $\begingroup$ What do you mean by "$G/G \to \mathbb{C}$"? Also, doesn't this confuse the representation ring $R(G)$ with the algebra $R(G) \otimes \mathbb{C}$? $\endgroup$
    – Smiley1000
    Commented Jun 22 at 20:33
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    $\begingroup$ @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. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 22 at 20:35
  • $\begingroup$ 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. $\endgroup$
    – Smiley1000
    Commented Jun 22 at 20:37
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    $\begingroup$ 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. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 22 at 20:38
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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).

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