Does $\mu_a\mu_b = (ag)(bg) = abg$ if the group $G$ hasn't been explicitly defined as abelian? Let $G$ be a group. I have a function $\mu:G \rightarrow G,\  \mu_a(g) = ag$ for some $a \in G$ and all $g \in G$. Then there's a function $\theta : G \rightarrow S_G, \ \theta(a) = \mu_a$. 
The claim is that $\theta$ is a homomorphism because $\theta(ab) = \mu_{ab} = \mu_a\mu_b$. I see that $\mu_{ab} = abg$, but can you necessarily say that $\mu_a\mu_b = (ag)(bg) = abg$ if the group hasn't been explicitly defined as abelian? I'd like to be able to pull the $g$ out, but I don't feel like I can do that. Does the group $G$ have to be abelian for me to pull the $g$ out?
 A: $\mu_a$ is a function from $G$ to $G$. It sends an element $g$ to $ag$.
The composition $\mu_a\circ\mu_b$ is then given by
$$\mu_a\circ\mu_b(g) = \mu_a(\mu_b(g)) = \mu_a(bg) = a(bg) = (ab)g = \mu_{ab}(g).$$
It is not equal to $(ag)(bg)$. 
Moreover, since $\mu_{a^{-1}}$ is the inverse of $\mu_a$, the latter is bijective.
That means that the map from $G$ to the set of bijective functions $\{f\colon G\to G\mid f\text{ is bijective}\}$, which is a group under composition of functions, given by $a\mapsto \mu_a$, is indeed a homomorphism. 
A: My answer hereunder doesn't add anything essential to previous valuable comments/accepted answer, but a way to represent the things (stacked maps) so as to reduce the risk of misinterpretation, as seemingly was in the OP.
Actually, your problem can be represented initially as:
\begin{alignat*}{2}
  \mu:G &\longrightarrow& X(G) \\
  a&\longmapsto& \mu_a:G &\longrightarrow G \\
  &&g&\longmapsto\mu_a(g):=ag \\
\tag 1
\end{alignat*}
where $X(G)$ is the set of the maps of $G$ in itself (we still don't know whether $\mu_a$ in a bijection for every $a \in G$). Now, $\forall a \in G, \mu_a(g)=\mu_a(h) \Rightarrow ag=ah \Rightarrow g=h$, so $\mu_a$ is injective for every $a \in G$. Besides, $\forall a,g \in G, g=\mu_a(a^{-1}g)$, so $\mu_a$ is also surjective for every $a \in G$. Conclusion: $\forall a \in G, \mu_a \in S_G$, and $(1)$ can be refined into:
\begin{alignat*}{2}
  \mu:G &\longrightarrow& S_G \\
  a&\longmapsto& \mu_a:G &\longrightarrow G \\
  &&g&\longmapsto\mu_a(g):=ag \\
\tag 2
\end{alignat*}
Finally, $a,b \in G \Rightarrow ab \in G \Rightarrow \mu_{ab} \in S_G$ (I think this latter membership is what you missed in the OP), so to state anything about $\mu_{ab}$, we have to let it work as map (bijection) on $G$. Now, for any given $a,b \in G$:
$$\forall g \in G, \mu_{ab}(g)=(ab)g=a(bg)=\mu_a(\mu_b(g))=(\mu_a\mu_b)(g) \Rightarrow \mu_{ab}=\mu_a\mu_b \tag 3$$
Your $\theta$ is $\mu$, and you are done.
As extra bonus, you can prove that $\mu$ itself is injective, so we have that $G$ embeds in $S_G$ (by left multiplication). This is Cayley's Theorem.
