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Suppose $G$ is a finite group and $V$ is a finite-dimensional representation of $G$ over a field $k$. I'd like to write $V$ as a tensor product $V_1 \otimes V_2 \otimes \dots V_n$ satisfying

(1) Each $V_i$ is a representation of $G$.

(2) No $V_i$ can be written as a tensor product of two or more $G$-representations of dimension > 1.

I'd like to call it a "prime factorization" of $V$, and each $V_i$, a prime representation of $V$?

Can this always be done? Is this prime factorization unique up to tensoring with one-dimensional representations (the units) and isomorphisms? Given $G$ and $k$ how can I write down all the prime representations of $G$ over $k$? What is a good reference for this stuff?

[Edited after R.'s comment.]

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  • $\begingroup$ In (2), you should probably say "two or more $G$-representations of dimension $>1$", as tensoring with $1$-dimensional representations is invertible. I see no reason to expect (3); for example if $G$ has a unique irreducible representation $V$ of dimension $>1$ (as is the case for $S_3$, $D_4$, or the quaternions, for example), then it seems to me that $V \otimes V$ has no other such factorisations. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Dec 29, 2020 at 19:01
  • $\begingroup$ @R.vanDobbendeBruyn, Of course! Thanks for pointing out the howlers. $\endgroup$
    – Peter
    Commented Dec 29, 2020 at 19:59
  • $\begingroup$ Doesn't the existence of such a tensor product decomoposition follow immediately by induction on the dimension of V? $\endgroup$
    – tj_
    Commented Dec 29, 2020 at 20:18
  • $\begingroup$ @tj_, Yes, that part is straightforward. $\endgroup$
    – Peter
    Commented Dec 29, 2020 at 20:21
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    $\begingroup$ Here is a counterexample to uniqueness: Take $G=\mathbb{Z}/2\mathbb{Z}$, with simple irreps denoted by $1$ and $-1$. Note that $(1\oplus -1)\otimes(1\oplus -1)\cong (1\oplus -1)\otimes(1\oplus 1).$ $\endgroup$
    – dhy
    Commented Dec 29, 2020 at 20:32

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These factorizations are not unique, and there is already a counterexample for the smallest nontrivial group $G = C_2$. The trivial representation has character $(1, 1)$ (where the first entry is the value of the character on the identity and the second entry is the value of the character on $-1$), and the sign representation has character $(1, -1)$, so representations can have character $(n + m, n - m)$ where $n, m$ are nonnegative integers. dhy in the comments gives the counterexample

$$(4, 0) = (2, 2) \times (2, 0) = (2, 0) \times (2, 0)$$

showing that we already don't have uniqueness here (any representation of prime dimension is prime). We can even arrange for the character values to be nonzero, for example

$$(35, 15) = (5, 5) \times (7, 3) = (5, 3) \times (7, 5).$$

Overall this is not a very well-behaved notion, I think; there are infinitely many representations of prime dimension given by taking direct sums of irreducibles whose dimension happens to add up to a prime. This is not a very natural condition and I don't think there's anything useful to say about it.

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  • $\begingroup$ This settles the unique factorization question. Is it useful to think about prime ideals in representation semi-rings at all? $\endgroup$
    – Peter
    Commented Dec 29, 2020 at 20:46
  • $\begingroup$ @Peter: ¯_(ツ)_/¯ As far as ideals go, there is the augmentation ideal which shows up in the Atiyah-Segal completion theorem: en.wikipedia.org/wiki/Atiyah%E2%80%93Segal_completion_theorem $\endgroup$
    – Qiaochu Yuan
    Commented Dec 29, 2020 at 20:49
  • $\begingroup$ There are situations where such a factorization does exist and is unique, as studied for example in Ferguson, Pamela A.; Turull, Alexandre. Prime characters and factorizations of quasiprimitive characters. Math. Z. 190 (1985), no. 4, 583--604. MR0808924. (The representations are over $k=\mathbb C.$) The theory of so-called $\pi$-special characters of $\pi$-separable groups (see Theorem D in that paper, for example) is important in the study of characters of solvable and related classes of groups. $\endgroup$
    – Tom WIlde
    Commented Apr 7, 2022 at 11:58

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