## Irreducible representations of finite abelian groups

Posted: February 8, 2011 in Representations of Finite Groups
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Theorem. A representation $\rho$ of a finite abelian group $G$ is irreducible if and only if $\deg \rho = 1.$

Proof.  One side of the theorem was already proved in Remark 2. Now, suppose that $\rho : G \longrightarrow \text{GL}(V)$ is an irreducible representation of $G.$ So, by definition, $V$ is a simple $\mathbb{C}[G]$-module and hence

$\text{End}_{\mathbb{C}[G]} V = \mathbb{C}, \ \ \ \ \ \ \ \ \ \ (*)$

by the second part of Diximier-Schur’s lemma . So if we show that $\rho(g)$ is a $\mathbb{C}[G]$-module homomorphism for all $g \in G,$ then by $(*),$ $\rho(g) \in \mathbb{C}$ for all $g \in G$ and hence $\deg \rho =1.$ Note that $\rho(g)$ is always a $\mathbb{C}$-module homomorphism but in order for $\rho(g)$ to be a $\mathbb{C}[G]$-module homomorphism as well, we need $G$ to be abelian, as you will see shortly. To show that $\rho(g)$ is a $\mathbb{C}[G]$-module endomorphism of $V,$ we need to prove that $\rho(g)(g'v)=g'\rho(g)(v),$ for all $g,g' \in G$ and all $v \in V.$ This is easy to see:

$\rho(g)(g'v)=\rho(g) \rho(g')(v)=\rho(gg')(v)=\rho(g'g)(v)=\rho(g')\rho(g)(v)=g'\rho(g)(v). \ \Box$

Corollary. The number of irreducible representations of a finite abelian group $G$ is $|G|.$

Proof. It is a trivial result of the above theorem and the lemma in this post. $\Box$

Question. Explain how to find all irreducible representations of a finite abelian group $G.$

Answer. First recall that, by the fundamental theorem of finite abelian groups, every finite abelian group is a finite direct product of cyclic groups. Also, we have already described all the representations of a finite cyclic group in Example 1. So, this is how we can find all $|G|$ irreducible (= degree one) representations of $G.$ We first write $G = \bigoplus_{j=1}^m \langle g_j \rangle,$ where $o(g_1)=n_1, \cdots , o(g_m)=n_m.$ Thus $|G|=n_1n_2 \cdots n_m.$ So every element of $g$ can be written uniquely as $g=g_1^{r_1}g_2^{r_2} \cdots g_m^{r_m}.$ Now let $\zeta_k, \ 1 \leq k \leq m,$ be an $n_k$-th root of unity. Define $\rho_{\zeta_1, \zeta_2, \cdots , \zeta_m} : G \longrightarrow \mathbb{C}^{\times}$ by

$\rho_{\zeta_1, \zeta_2, \cdots , \zeta_m}(g_1^{r_1} g_2^{r_2} \cdots g_m^{r_m}) = \zeta_1^{r_1} \zeta_2^{r_2} \cdots \zeta_m^{r_m},$

for all $0 \leq r_k \leq n_k-1, \ 1 \leq k \leq m.$ Clearly each $\rho_{\zeta_1, \zeta_2, \cdots , \zeta_m}$ is a group homomorphism and hence a representation of degree one for $G.$ The number of these representations is $|G|$ because each $\zeta_k$ has $n_k$ possible values and thus we have $n_1n_2 \cdots n_m=n$ possible representations of this form. Also, by the Corollary, we know that $G$ has exactly $n$ irreducible representations. So $\rho_{\zeta_1, \zeta_2, \cdots , \zeta_m}$ are all the irreducible representations of $G. \ \Box$

Example. Find all irreducible representations of $K_4,$ the Klein-four group.

Solution. $K_4$ is the direct product of two cyclic groups of order $2.$ So $n_1=n_2=2.$ Let $g_1, g_2$ be the generators of these groups. Then $K_4=\{1, g_1,g_2,g_1g_2 \}.$ Now the second roots of unity are $\pm 1.$ Thus the irreducible representations of $K_4$ are defined by

1) $\rho_{1,1}(g_1^ig_2^j)=(1)^i(1)^j=1,$

2) $\rho_{1,-1}(g_1^ig_2^j)=(1)^i(-1)^j=(-1)^j,$

3) $\rho_{-1,1}(g_1^ig_2^j)=(-1)^i(1)^j=(-1)^i,$

4) $\rho_{-1,-1}(g_1^ig_2^j)=(-1)^i(-1)^j=(-1)^{i+j},$

where $0 \leq i,j \leq 1. \ \Box$