## 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. $\Box$