Is Hom$(G_1, G_2)$ a group? The collection of all homomorphisms from the group $G_1$ to the group $G_2$ is denoted as Hom$(G_1, G_2)$. I am willing to show that if $G_1 \simeq  G_1'$ then Hom$(G_1, G_2) \simeq$  Hom$(G1', G2)$. But the problem is I am unable to find out whether Hom$(G_1, G_2)$ is a group or not. If yes, under what binary operation will it be?
Any suggestions will be highly appreciated. thanks in advance
 A: As Mike pointed out, $\operatorname{Hom}(G,H)$ is not in general a group. However, it is a group if $H$ is abelian, and this observation can lead you to a rabbit hole.
It has a category-theoretical intepretation. Suppose we have a category $\mathcal C$ satisfying some natural assumptions (existence of finite products). Then an object $G$ in $\mathcal C$ is called a group object in $\mathcal C$ (or just a group in $\mathcal C$) if we have a natural group structure on $\operatorname{Hom}(X,G)$, for any object $X$ of $\mathcal C$.
It turns out that groups in the category of groups are exactly the abelian groups, which may seem kind of funny. On the other hand, groups in the category of sets are your regular run-of-the-mill groups, while groups in the category of topological spaces and algebraic varieties, for example, are what you would expect: topological and algebraic groups, respectively. Similarly for smooth manifolds and Lie groups.
There is also a dual concept of cogroup (i.e. $G$ such that $\operatorname{Hom}(G,X)$ has a natural group structure), and a closely related concept of Hopf algebras, which are found in several branches of mathematics (for example, here in Wrocław, they are studied independently by analysts and algebraists).
A: There is, in general, no natural group structure on $\text{Hom}(G_1, G_2)$. If $G_2$ is an abelian group there is one: if $\varphi, \psi \in \text{Hom}(G_1, G_2)$, define $(\varphi+\psi)(x)=\varphi(x)+\psi(x)$. In verifying that this is indeed an (abelian!) group, you will likely see what fails when $G_2$ is not abelian.
$\text{Hom}(G_1, G_2)$ is of course a set, and indeed there is a natural bijection between $\text{Hom}(G_1, G_2)$ and $\text{Hom}(G_1', G_2)$ - perhaps you'd like to prove that.
Additionally, there is a natural group structure on the set of isomorphisms from a group $G$ to itself; this is denoted $\text{Aut}(G)$. The group structure here, instead of just the group operation as above, is composition: $(\varphi \circ \psi)(x) = \varphi(\psi(x))$, and our identity is the identity map. Again, this is worth convincing yourself of. (Thanks to MikeF for pointing out that I mistakenly wrote that $\text{Hom}(G,G)$ has a group structure in this manner. It doesn't - the zero homomorphism can't be invertible!)
