Prove a mapping is a normal subgroup

Given a problem like this... Let @:G-->H be a homomorphism of G onto H, and let N be a normal subgroup of G. Show that @(N) is a normal subgroup of H. How do I prove that a mapping is a normal subgroup of a group? What I am missing here is some understanding of the terminology and some clear understanding of mappings, homomorphisms, subgroups and normal subgroups. Please attempt to give me some understanding of what a problem like this is asking me. How can a mapping be a normal subgroup? What is a mapping from G-->H ? Is the mapping a set or group? Any help you can give would be great!

Subgroups and indexes

Explain what the index of a subgroup and a coset of a group are. Also, prove that if N is a subgroup of a group G such that [G: N] = 2, and if "a" and "b" are elements of G, then the product "ab" is an element of N if and only if either (1) both "a" and "b" are elements of N or (2) neither "a" nor "b" is an element of N.

Conjugacy of group elements

Show that conjugacy of group elements is an equivalence relation.

Conjugacy Classes

What are the conjugacy classes in S_3?

Equivalence relations, surjective maps, partitions and fibers??

Consider any surjective map f from a set X onto another set Y. We can define a relation on X by x_1 ~ x_2 if f(x_1) = f(x_2). Check that this is an equivalence relation. Show that the associated partition of X is the partition into "fibers" f^(-1) (y) for y in Y. I would like to understand what this question is asking me and how I should answer such a question. There are many questions like this one and I would like to learn enough from your solution and explanation of this one to enable me to possibly be able to attempt to answer similar ones on my own. As much information as possible would be nice.