Problem in finding a bijective map. In my book it has been claimed that there is a one-to-one correspondence between the set of all normal subgroups of a group and the set of all homomorphisms defined on it.But I have failed to find out such map.I know that every homomorphism defined on a group yields a normal subgroup which is precisely the kernel of the homomorphism and conversely every normal subgroup of a group can be thought of as the kernel of the associated natural homomorphism. Now how can I proceed? Please help me.
Thank you in advance.
 A: As was noted in the comments, the result is false as stated. Indeed, if $G$ is any group, $h:G\to H$ is any surjective homomorphism, and $X$ is any set such that $|X|=|H|$, let $i:H\to X$ be a bijection. We can define a group operation on $X$ in such a way that $i$ is an isomorphism, and $i\circ h$ is then a homomorphism of $G$ onto $X$ with the same kernel as $h$. Thus, for each normal $N\unlhd G$ the collection of surjective homomorphisms with kernel $N$ is not just infinite: it’s a proper class. I suspect that something like the following was intended.
Suppose that $h_0:G\to H_0$ and $h_1:G\to H_1$ are homomorphisms. Write $h_0\sim h_1$ if and only if there is an isomorphism $i:H_0\to H_1$ such that $h_1=i\circ h_0$. It’s easy to check that $h_0\sim h_1$ if and only if $\ker h_0=\ker h_1$. It’s not hard to check that if $\mathscr{H}$ is any set of homomorphisms with domain $G$, and we restrict the definition of $\sim$ to members of $\mathscr{H}$, then $\sim$ is an equivalence relation on $\mathscr{H}$. Without that restriction $\sim$ isn’t even a set, but we can think of it as a proper class that is an equivalence relation on the proper class of homomorphisms with domain $G$. 
Its equivalence classes are also proper classes, but each of these equivalence classes has a distinguished member: for each $N\unlhd G$ we have a quotient group $G/N$ and an associated quotient map $q_N:G\to G/N$, and if $h:G\to H$ is a surjective homomorphism, then $h\sim q_{\ker h}$. If $Q=\{q_N:N\unlhd G\}$, then $Q$ contains exactly one member of each equivalence class, and there is certainly a bijection between $Q$ and the set of normal subgroups of $G$.
In short, there is a bijection between normal subgroups of $G$ and quotient maps on $G$, and each homomorphism with domain $G$ corresponds in a natural way to a unique quotient map on $G$. To correct the statement in the book, replace homomorphisms by quotient maps, and then note that every homomorphism on $G$ corresponds naturally to a unique quotient map on $G$.
