The natural projection mapping $\pi : G \to G/N$ defined by $\pi(x) = xN$, for all $x$ in $G$, is a homomorphism, and $\ker(\pi) = N$. The question again...

The natural projection mapping $\pi : G \to G/N$ defined by $\pi(x) = xN$, for all $x$ in $G$, is a homomorphism, and $\ker(\pi) = N$.

I am wanting to prove that $\ker(\pi) = N$, where $N$ is a normal subgroup of $G$, but I have been trying to think of a way to begin but I cannot seem to get anywhere despite my multiple efforts.
Thank you in advance.
 A: Given a subgroup $H\leqslant G$ and an element in $a\in G$, we have the left coset
$$aH = \{ah :h\in H\}, $$
and similarly the right coset
$$Ha = \{ha :h\in h\}. $$
The distinction, of course, is in the order of multiplication. A normal subgroup $N\mathrel{\unlhd}G$ is a subgroup of $G$ with the property that
$$gN = Ng \tag 1$$
for all $g\in G$. Another way of stating this is that $N$ is stable under conjugation, that is if $n\in N$ and $g\in G$, then $gng^{-1}\in N$ - this is the conjugation of $n$ by $g$. To see this, right-multiplying $(1)$ yields
$$gNg^{-1} = N.$$
Now, to the actual question - the map $\pi:G\to G/N$ with $g\mapsto gN$ is simply the map from $g$ to the left coset of $N$ in $g$. But if $N$ is a normal subgroup, then $gN=Ng$, and hence for $a,b\in G$
$$\pi(a)\pi(b) = (aN)(bN) = (aN)(Nb) = aNb = (aN)b = (ab)N = \pi(ab).$$
Finally, if $a\in\operatorname{Ker}(\pi)$ then $\pi(a)=N$, which is is the same as $aN=N$, which implies that $a\in N$.
A: Subgroups are closed under the group operation.  If $a\in N$, what is the set $aN=\{an|n\in N\}?$  What is the identity element of $G/N$?  This will give you $N\subseteq\operatorname{ker}(\pi)$.  The reverse implication is quite similar, can you show it?
A: This is a named theorem, called the "first isomorphism theorem". So if you want to see a proof simply open your favorite text on group theory or algebra, turn to the index and look it up. It is such an important theorem that it will be there!
It is so important because it says that every normal subgroup corresponds to the kernel of a homomorphism, and of course we know that the kernel of every homomorphism is a normal subgroup. So, normal subgroups and homomorphisms are dual to one another. so, for example, a simple group is a group which contains no non-trivial, proper normal subgroups, so we now know that an equivalent definition of a simple group is one which contains no homomorphic images apart from isomorphisms and the trivial map (so, collapsing everything to the identity).
