Identity and Inverse Homomorphisms

Question

For a group G and an abelian group H, Hom(G,H) is the set of all homomorphisms from G to H. My notes from class talk about the identity and the inverse homomorphism- I was wondering what these are?

Thanks.

Answer 1

When $G$ is a group and $H$ is an (additive) abelian group, $\hom_{\text{grp}} (G,H)$ forms an (additive) abelian group with pointwise addition: $(f_1+f_2)(g)=f_1(g)+f_2(g)$, for $f_1,f_2\in \hom (G,H)$ and $g\in G$. You should check that the sum of two homomorphisms is a homomorphism. The identity in this group is the zero homomorphism, while the inverse of a homomorphism is given by $(-f)(g)=-f(g)$, where $f\in \hom (G,H)$ and $g\in G$.

Answer 2

The identity map $1_G:G \rightarrow G$ on a set is the function which sends each element to itself.

You should verify that this is a homomorphism, and that when a homomorphism is a bijection, the inverse function is also a homomorphism.

Answer 3

If $(H,+)$ is an abelian group, then $Hom (G,H)$ can be given a group structure. Infact, given two homomorphisms $f,g: G \longrightarrow H$ we can define their sum $f+g : G \longrightarrow H$ componentwise with the formula $$(f+g)(x) = f(x) + g(x)$$

The identity of $Hom (G,H)$ is just the $0$ homomorphism, and the inverse of $f$ is just defined by $(-f)(x) = -(f(x))$.