# Definitions of $\mathrm{Hom}(V,W)$

I have the definition of a homomorphism as map such that $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)$

I have the definition of $\mathrm{Hom}(V,W)$ as

\begin{align}\mathrm{Hom}(V,W) &= \{\mathbb{C}-\text{linear maps }\varphi:V \rightarrow W \} \\ &\cong \{n \times m \text{ matrices } \} \end{align}

Does $\mathrm{Hom}$ stand for homomorphisms because I cannot see how it relates to the definition of homomorphism

I have the definition of $\mathrm{Hom}_G$ as

\begin{align}\mathrm{Hom}_G(V,W) &= \{\varphi \in \mathrm{Hom}(V,W) \mid g\varphi(v)=\varphi(gv), \forall g \in G, \forall v \in V\} \\ &=\{\varphi \in \mathrm{Hom}(V,W) \mid \rho(g)\cdot\varphi(v)=\varphi(\rho(g)v) , \forall g \in G, \forall v \in V\} \end{align}

I then have the question:

Prove that if $\rho$ is an irreducible representation, then for an element $g \in G$

$$g \in Z(G) \iff \rho(g)=\lambda I$$

In the solution I have that:

$(\implies")$ If $g \in Z(G)$ then $gh=hg \ \forall h \in G$. By definition of $\mathrm{Hom}_G$ this means that $\rho(g) \in \mathrm{Hom}_G(V,V)$. But \rho is irreducible so $\mathrm{Hom}_G(V,V)$ consists of scalar matrices by Schur's Lemma.

I do not understand how "By definition of $\mathrm{Hom}_G$ this means that $\rho(g) \in \mathrm{Hom}_G(V,V)$. " How is this the definiton of $\mathrm{Hom}_G$? I cannot see why this is equivalent.

• Why do you say that $\rho(g) \rho(h)= \rho(h) \rho(g)$ ? In general $G$ need not to be commutative. – Crostul May 30 '15 at 11:01
• There should not be an equal sign between the set of linear maps and the set of matrices; these spaces are merely isomorphic – Hagen von Eitzen May 30 '15 at 11:09
• I was trying to quote a small part of the solutions. I will edit the question to make things clearer. – Permian May 31 '15 at 10:52
• What about the trivial representation? In that case, $\rho(g)=I$ for all $g\in G$, so the statement is false if $G$ is not abelian. – Pierre-Guy Plamondon May 31 '15 at 21:19
• That is weird because doesnt state it needs to be non abelian – Permian Jun 1 '15 at 9:00

For your second question, assume that $g$ is an element of the center of $G$, and let $\varphi = \rho(g)$. Let $h$ be an element of $G$. Since $g$ is in the center, we have $gh=hg$, so $\rho(g)\cdot \rho(h) = \rho(h)\cdot\rho(g)$. By our notation, this can be written as $\rho(h)\cdot\varphi = \varphi\cdot\rho(h)$; in other words, for any $v\in V$, we have that $\rho(h)\cdot\varphi(v) = \varphi(\rho(h)(v))$. By definition of $Hom_G$, this means that $\varphi\in Hom_G(V,V)$. Since $\varphi=\rho(g)$, this answers your question.
• I still dont understand your second paragraph. I have the definition for $\mathrm{Hom}_G$ as (crudely) $g\varphi=\varphi g$ which doesnt mean $\rho(g)\rho(h)=\rho(h)\rho(g)$. How is this rectified???? – Permian Jun 1 '15 at 9:03