## Number of non-equivalent irreducible representations (2)

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