I'm a physicist and struggling with a question, kinda lost and would really appreciate some help. I'm supposed to find all irreducible, nonisomorphic representations of $\mathbb{Z}_3\times \mathbb{Z}_4$.

I'm familiar with two things:

  1. If we have two groups: $G_1 \rightarrow GL(V_1) \text{ and } G_2 \rightarrow GL(V_2)$. Every irreducible representation of $G_1\times G_2$ will be of the form $\rho_1 \otimes\rho_2$, where $\rho_1$ and $\rho_2$ are irreducible representations of $G_1 $ and $G_2$ respectively.

  2. The squared dimension of all irreducible representations (irrps or irreps) should amount to the order of the group $|G|$ : $\sum_{i=\text{irrps}} \dim(V_i)^2=|G|.$

My plan is to find all irreducible representations of $\mathbb{Z}_3$ and $ \mathbb{Z}_4$ respectively and then form all posible tensor products to get all nonisomorphic irreducible representations of $\mathbb{Z}_3\times \mathbb{Z}_4$.

For $\mathbb{Z}_3$ we have the trivial representation: $$\rho_{trivial}: \qquad[0],[1],[2]\rightarrow 1 ,$$

we have the reducible 2D representation:

$$[0]\rightarrow \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}, \qquad [1] \rightarrow \begin{pmatrix} 0 & -1\\ 1 & -1 \end{pmatrix} \qquad [2] \rightarrow \begin{pmatrix} -1 & 1\\ -1& 0 \end{pmatrix},$$

which splits into two 1D irr. representations: $$\qquad \rho_1: \qquad [0]\rightarrow 1, \qquad [1]\rightarrow e^{\frac{2}{3}\pi i}, \qquad [2] \rightarrow -e^{\frac{1}{3}\pi i}$$


$$\qquad \rho_2: \qquad [0]\rightarrow 1, \qquad [1] \rightarrow -e^{\frac{1}{3}\pi i}, \qquad [2] \rightarrow e^{\frac{2}{3}\pi i} .$$

The dimensions add up: $\dim(\rho_{trivial})^2+\dim(\rho_{1})^2+\dim(\rho_{2})^2 =1+1+1=3=|G| .$

For $\mathbb{Z}_4:$

$$[0] \rightarrow \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}, [1] \rightarrow \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}, [2]\rightarrow \begin{pmatrix} -1 & 0\\ 0 & 1 \end{pmatrix}, [3] \rightarrow \begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix},$$ gives two irr. representations $\rho_1,\rho_2$:

$$\rho_1:\qquad [0],[1],[2],[3]\rightarrow 1,1,-1,-1 \qquad \text{(respectively)}$$ $$\rho_2:\qquad [0],[1],[2],[3]\rightarrow 1,-1,1,-1 \qquad \text{(respectively)}.$$

If my thinking is correct there should be 2 more 1D irr. representations, so that the dimensions add up.


Is my thinking here regarding the tensor product and the irreps of $\mathbb{Z}_3$ and $\mathbb{Z}_4$ correct? If my thinking is correct, the set of all $\mathbb{Z}_3\times \mathbb{Z}_4$ irreps should consist of 12 1D irreducible representations?

See also: Related


1 Answer 1


You are correct with the number of representations, but you should consider that there is an easier solution. If you know the representations of an arbitrary cyclic group, then note that $\mathbb Z_3\times \mathbb Z_4\cong \mathbb Z_{12}$, the cyclic group of order $12$, and you are done. Whenever $m$ and $n$ are relatively prime, we have that $\mathbb Z_m\times\mathbb Z_n\cong \mathbb Z_{mn}$.

I'm thinking that also perhaps you are confusing $\mathbb Z_4$ with $\mathbb Z_2\times \mathbb Z_2$. The function you describe for $\mathbb Z_4$ into the general linear group gives a subgroup isomorphic to $\mathbb Z_2\times\mathbb Z_2$ and cannot be a homomorphism from $\mathbb Z_4$.

  • $\begingroup$ So by taking the tensor product of all $\mathbb{Z}_3$ and $\mathbb{Z}_4$ irreducible representations I should get all the irr. representations of $\mathbb{Z}_3 \times \mathbb{Z}_4$? If I understand your answer correctly my two irreducible representations for $\mathbb{Z}_4$ are not correct? I mainly don't understand the second part of your answer. $\endgroup$
    – Luka
    May 3, 2020 at 18:28
  • $\begingroup$ @Luka Your representations of $\mathbb Z_4$ are not correct, yes. $\endgroup$ May 3, 2020 at 18:36
  • $\begingroup$ Many thanks! So $\mathbb{Z}_3\times \mathbb{Z}_4$ is isomorphic to $\mathbb{Z}_{12}$. And $\mathbb{Z}_{12}\cong C_{12}$, so by mapping all the $[n]$ (in $C_{12}$ or $Z_{12}$) elements to $n$ roots of unity in all the various ways, we get all the different irreducible representations of $C_{12}$ or $Z_{12}$? I apologize if this sound like a stupid question, like I've said I'm not really an expert in abstract algebra. $\endgroup$
    – Luka
    May 3, 2020 at 19:14
  • $\begingroup$ @Luka Right. You only have to worry about where you send a generator. $\endgroup$ May 3, 2020 at 19:16
  • $\begingroup$ Oh I get it now, many thanks!! $\endgroup$
    – Luka
    May 3, 2020 at 19:23

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