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

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