Number of non-equivalent irreducible representations (2)

Posted: February 10, 2011 in Representations of Finite Groups
Tags: , , ,

I am now going to look at some consequences of the theorem we proved in part (1).

1) Let G be a finite abelian group. Then, by the corollary in this post, the number of irreducible representations of G is |G| and all such representations have degree one by the theorem in that post. The theorem in part (1) gives the same result: since G is abelian, the number of conjugacy classes of G is |G|. Also, since \mathbb{C}[G] is abelian, in the identity (2) in the proof of that theorem, we must have n_1=n_2= \cdots = n_r=1.

2) A finite group G is abelian if and only if every irreducible representation of G has degree one.

Proof.  Well, G is abelian if and only if \mathbb{C}[G] is abelian. Now (2) in the proof of the theorem in part (1) shows that \mathbb{C}[G] is abelian if and only if n_1=n_2 = \cdots = n_r = 1. \Box

3) The number of (pairwise non-equivalent) irreducible representations of S_3 is 3. One of these representations has degree two and the rest have degree one.

Proof. By the theorem in part (1), we have 6=|S_3|=\sum_{i=1}^r n_i^2. We also saw in the second remark in here that the number of degree one representations of S_3 is two.  So n_1=n_2=1 and hence the only possibility now is that r=3 and n_3=2. This also proves the algebra isomorphism

\mathbb{C}[S_3] \cong \mathbb{C} \times \mathbb{C} \times M_2(\mathbb{C}). \Box

4) The number of (pairwise non-equivalent) irreducible representations of quaternion group Q_8 is 5. One of these representations has degree two and the rest have degree one.

Proof.  Again by the theorem in part (1), we have 8=|Q_8|=\sum_{i=1}^r n_i^2. We also proved in the third remark in here that Q_8 has exactly four representations of degree one. So n_1=n_2=n_3=n_4=1. Clearly, the only possibility now is that r=5 and n_5=2. This also proves the algebra isomorphism

\mathbb{C}[Q_8] \cong \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times M_2(\mathbb{C}). \Box

5) The number of (pairwise non-equivalent) irreducible representations of the dihedral group of order eight D_8 is 5. One of these representations has degree two and the rest have degree one.

Proof. See the fourth remark in here and copy the above proof for Q_8. We also get

\mathbb{C}[D_8] \cong \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times \mathbb{C} \times M_2(\mathbb{C}). \Box

Remark. By 4) and 5), \mathbb{C}[D_8] \cong \mathbb{C}[Q_8], although D_8 \ncong Q_8. See also here.

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