Number of representations of degree one (1)

Posted: February 8, 2011 in Representations of Finite Groups
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Lemma. Let G be a finite abelian group. The number of degree one representations of G is |G|.

Proof. Let |G|=n. By the fundamental theorem for finite abelian groups, G = C_1 \times C_2 \times \cdots \times C_k, where each C_j is a cyclic group of order, say, n_j. Clearly n_1n_2 \cdots n_k = n. So every element of G is written uniquely as g_1g_2 \cdots g_k, where g_j \in C_j. By Example 1 each C_j has exactly n_j degree one representations. Let R_j be the set of degree one representations of C_j and let R be the set of degree one representations of G. Define f : R \longrightarrow R_1 \times R_2 \times \cdots \times R_k by

f(\rho)=(\rho_1, \rho_2, \cdots , \rho_k),

where \rho_j = \rho |_{C_j}. If we show that f is a bijection, then |R|=\prod |R_j|=\prod n_j = n and the lemma is proved.

1) f is well-defined : This is obviouse because if \rho \in R, then clearly \rho_j \in R_j.

2) f is injective : If \rho, \mu \in R and \rho_j = \mu_j, for all j, then for any g = g_1g_2 \cdots g_k \in G we have

\rho(g)=\prod \rho(g_j) = \prod \rho_j(g_j) = \prod \mu_j(g_j)= \prod \mu(g_j)=\mu(g).

Thus \rho = \mu and so f is injective.

3) f is surjective : Let \mu_j \in R_j, \ 1 \leq j \leq k and define \rho : G \longrightarrow \mathbb{C}^{\times} by

\rho (g_1g_2 \cdots g_k)=\mu_1(g_1) \mu_2(g_2) \cdots \mu_k(g_k),

for all g_j \in C_j. It is easy to see that \rho \in R. Also \rho(g_j)=\mu_j(g_j) for all g_j \in C_j, i.e. \mu_j = \rho|_{C_j}=\rho_j and thus f(\rho)=(\mu_1, \mu_2, \cdots , \mu_k). \ \Box

Theorem. Let G be a finite group. The number of degree one representations of G is [G:G'].

Proof. Since G/G' is a finite abelian group, the number of degree one representations of G/G' is |G/G'|=[G:G'], by the above lemma. So we only need to define a bijection between R, the set of degree one representations of G and S, the set of degree one representations of G/G'. Let \pi : G \longrightarrow G/G' be the natural homomorphism and define f : S \longrightarrow R by

f(\rho) = \rho \pi.

It is immediate that f is well-defined and injective. So we only need to show that f is surjective. To see this, let \bar{\rho} \in R. So \bar{\rho} : G \longrightarrow \mathbb{C}^{\times} is a group homomorphism and hence G/\ker \bar{\rho} is abelian because it is isomorphic to a subgroup of \mathbb{C}^{\times}. Thus

G' \subseteq \ker \bar{\rho}. \ \ \ \ \ \ \ \ \ \ \ \ (*)

Now define \rho : G/G' \longrightarrow \mathbb{C}^{\times} by

\rho(gG')=\bar{\rho}(g),

for all g \in G. If g \in G', then g \in \ker \bar{\rho}, by (*), and hence \bar{\rho}(g)=1. Thus \rho is well-defined. Clearly \rho is a group homomorphism because \bar{\rho} is so. Thus \rho \in S. Finally f(\rho)(g)=\rho \pi(g)=\rho(gG')=\bar{\rho}(g), for all g \in G. Hence f(\rho) = \bar{\rho} and the proof is complete. \Box

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