Show that $Hom(G,GL_m(\mathbb{C}))$ is not a group with point-wise multiplication Suppose A as a finite abelian group, then I know $\hat{A}:=Hom(A,\mathbb{C}^{\times})$ is a group with operation $\rho_1\rho_2(g)=\rho_1(g)\rho_2(g)$ for $\rho_1,\rho_2\in \hat{A}, g\in G$. However, for the set of higher dimensional linear representations it's not a group. Is it because the inverse of $a\in\hat{A}$ is from $GL_{m}(\mathbb{C})$ to $A$ but $GL_m(\mathbb{C})$ is not abelian? Thanks in advance!
Edit: as paul blart math cop mentioned, I correct $G$ to $A$.
 A: $\newcommand{\grp}{\mathsf{Grp}}$
$\newcommand{\set}{\mathsf{Set}}$
Note that if we were instead looking at the collection of set functions $\set(A, GL_m(\mathbb C))$ under pointwise operations then this is a group. In fact, this only depends on the underlying set of $A$. So the question becomes why the set of all group homomorphisms $\hat{A} = \grp(A, GL_m(\mathbb C))$ is not a subgroup of $\set(A, GL_m(\mathbb C))$ for $m > 1$. The two ways this can fail are closure under multiplication and closure under inversion. I'll touch on both.
Suppose we had $\rho_1, \rho_2 \in \grp(A, GL_m(\mathbb C))$. To have $\rho_1 \rho_2 \in \grp(A, GL_m(\mathbb C))$ would mean that the function $\rho_1 \rho_2: A \longrightarrow GL_m(\mathbb C)$ is a group homomorphism. That is, $(\rho_1 \rho_2)(ab) = (\rho_1 \rho_2)(a) (\rho_1 \rho_2)(b)$. The left hand side reduces to $\rho_1(ab) \rho_2(ab) = \rho_1(a) \rho_1(b) \rho_2(a) \rho_2(b)$. The right hand side reduces to $\rho_1(a) \rho_2(a) \rho_1(b) \rho_2(b)$. Hence, closure under multiplication of $\grp(A, GL_m(\mathbb C))$ comes down to the equation
$$
\rho_1(a) \rho_1(b) \rho_2(a) \rho_2(b) = \rho_1(a) \rho_2(a) \rho_1(b) \rho_2(b)
$$
for all $a, b \in A$ and $\rho_1, \rho_2 \in \grp(A, GL_m(\mathbb C))$. This holds precisely when $\rho_1(b)$ and $\rho_2(a)$ commute. That is, we will have $\rho_1 \rho_2 \in \grp(A, GL_m(\mathbb C))$ if and only if $\rho_1(a)$ and $\rho_2(b)$ commute for every possible $a, b \in A$. This is very unlikely to hold, since $GL_m(\mathbb C)$ is not abelian. It is not completely impossible, as it does hold for $A = 0$ the trivial group. But for an explicit counterexample, take $A = \mathbb Z/2$, $\rho_1(1) = \pmatrix{1 & 1 \\ 0 & -1}$ and $\rho_2(1) = \pmatrix{1 & 2 \\ 0 & -1}$. These are well defined group homomorphisms but $\rho_1(1)$ and $\rho_2(1)$ do not commute so their product $\rho_1 \rho_2: \mathbb Z/2 \longrightarrow GL_m(\mathbb C)$ is not a group homomorphism.
The same sort of issue actually cannot happen with inversion. If $\rho \in \grp(A, GL_m(\mathbb C))$ we will have $\rho^{-1} \in \grp(A, GL_m(\mathbb C))$. Indeed, $\rho^{-1}$ is defined as $a \mapsto (\rho(a))^{-1}$. The equation we're looking for is $\rho^{-1}(ab) = \rho^{-1}(a) \rho^{-1}(b)$. The left hand side is $\rho^{-1}(ab) = (\rho(a) \rho(b))^{-1} = \rho(b)^{-1} \rho(a)^{-1}$. On the other hand, the right hand side is $\rho^{-1}(a) \rho^{-1}(b) = \rho(a)^{-1} \rho(b)^{-1}$. Equating these two expressions yields
$$
\begin{align*}
\rho^{-1}(ab) &= \rho^{-1}(a) \rho^{-1}(b)\\
&\Updownarrow\\
\rho(b)^{-1} \rho(a)^{-1} &= \rho(a)^{-1} \rho(b)^{-1}\\
&\Updownarrow\\
\rho(a) \rho(b) &= \rho(b) \rho(a).
\end{align*}
$$
This equality always holds because $A$ is abelian. Indeed, the left hand side of the last equation is $\rho(ab)$ and the right hand side is $\rho(ba)$, which are equal as $ab = ba$ in $A$.
