Is there a way to symbolically denote a Homomorphism?

Question

The First Isomorphism Theorem is (I think) the first to explicitly Declare a group homomorphism:

Given groups $(G,*),(H,\circ)$ (Or just $G,H$). If $f: G \rightarrow H$ is a homomorphism then:

1. $\ker(f) \triangleleft G$
2. $G / \ker(f) \cong \operatorname{im}(f) \leq H$

There is quite a lot of information contained in the few relations encoded in this first consequence (and the proof) of the property of being a group homomorphism.

My main question:

Given $G,H$: Does the collection of all homomorphisms $G \to H$ qualify as a set, generally speaking? In any case, is there an abbreviation?

Related secondary questions:

1. There exist a homomorphism with domain $G$ and kernel $K \triangleleft G$ for each $K \triangleleft G$. Does every $K$ correspond to a target group (up to isomorphism)?

2. Do all groups isomorphic to $G/K$ correspond to a homomorphism with domain $G$?

3. What is the valid way (if any) of stating "$f \in Grp(G,H)$" ?

Answer 1

$0)$ The collection of all set functions $G\to H$ is a set (since $G$ and $H$ are both sets), and since the collection of all homomorphisms $G\to H$ is just a subset of this, it is certainly a set as well. From the point of view of category theory, if we let $\bf{Grp}$ denote the category of groups, then you can denote this set by $\hom_{\bf{Grp}}(G,H)$, or $\operatorname{Mor}_{\bf{Grp}}(G,H)$. If the category is clear from context we'd usually just write $\hom(G,H)$.

$1)$ Yes. If $K$ is a normal subgroup of $G$, then $G/K$ is a group, and $K$ is the kernel of the "natural" projection homomorphism $G\to G/K$, which sends $g\mapsto gK$.

$2)$ If $H$ is isomorphic to $G/K$, then you can choose an isomorphism $\varphi:G/K\to H$, and take the composition

$$G\to G/K\overset{\varphi}{\to}H,$$

where the first map is again the natural projection, to get a homomorphism $G\to H$.

$3)$ Yes, going back to "$0$", we can just write $f\in\hom_{\bf{Grp}}(G,H)$, or $f\in\hom(G,H)$.