Normal Groups and Quotient Groups These concepts are currently confusing me. My reading first defined a normal subgroup as one that is the kernel of a group homomorphism. Then it introduced the terms "left coset" and "right coset," which seem straightforward enough. 
The next definition was that of the index of a subgroup $[G:H]$. Then, it gave another definition of a normal subgroup (?), namely, one invariant under conjugation ($gNg^{-1} = N \,\,\forall \, g \in G$). 
Finally, it gave the following tortuous definition of a quotient group: The set of all left cosets of a normal subgroup $N$ with the law of composition $(gN)(hN) = (gh)N$ and having order $[G:N]$.
Why are those definitions equivalent? And what does that last horrendous definition mean? I'd really appreciate it if someone could provide an intuitive/simple explanation of these terms.  
 A: The elements of $G/N$ (where this indicates the quotient group) are cosets of $N$. So, if I reach into $G/N$ and pull out an element, that element has the form $gN$, where $g$ is in $G$. 
The group operation of $G/N$ is written as multiplication usually. So, when I multiply two elements of $G/N$, say $gN$ and $hN$, I have $(gN) \cdot (hN) = (gh)N$.   This defines how we multiply in the quotient group $G/N$. 
Here are some thoughts on the equivalence of the two "normal" definitions:
Let $\phi$ be a group homomorphism and $N$ its kernel. So by definition 1, $N$ is normal in $G$. Take any element of $G$, call it $g$. Let $n$ be any element of $\ker(\phi) = N$. Then $$\phi(g^{-1}ng) =\phi(g^{-1})\phi(n)\phi(g) = \phi(g^{-1})1\phi(g) = \phi(g^{-1}g) = \phi(1) = 1$$
So anything in $g^{-1}Ng$ is also in $N$ (as $N$ is the kernel of $\phi$). 
A: Have you seen the motivation of cosets as an equivalence classes? If we have $H 
\leq G$ then say that $a \sim b$ if $a^{-1}b \in H$ (check this is an equivalence if you've never seen this before). It turns out this holds if and only if $aH = bH$. Note that this is exactly what we do in modular arithmetic where $a = b \mod n$ iff $a-b \in n\mathbb{Z}$. So what we are morally doing is just forgetting the component of $a$ that comes from $H$ and only caring about the "residue". Now there are a few problems: to make a meaninful group operation out of this we'd like residues to not depend on order, i.e. here we're allowing ourselves to forget components from $H$ on the right, but in a non-Abelian group this may be different from what happens when we forget factors from $H$ on the left. 
The other problem is that in a general group we have no structure like the order on the integers that lets us just write a coset as a "privileged" element. So while in modular arithmetic we can get away with writing $\mathbb{Z}/n$ as $\{0,1,...,n-1\}$ what we really meant was $\{0 + n\mathbb{Z},...,(n-1)+n\mathbb{Z}\}$ the cosets. So when we define quotients in general we have to just leave the cosets as cosets. 
A: I share your confusion. I once found the definition of normal subgroup unnatural. Why $g N g^{-1} = N$, right?
This is how I usually prefer to see normal subgroups: $N$ is normal if and only if the operation on cosets of $N$ is closed. This also gives me a better intuition why a normal subgroup is precisely a kernel of some homomorphism.
Here's a proof that the alternative definition is equivalent to the original definition.
Given two cosets $gN$ and $hN$, when will $(gN)(hN)$ be another coset of $N$? Well, if $N$ is normal, we can write
$$
gNhN = g(hNh^{-1})hN = ghNN = ghN.
$$
This proves that $N$ being normal is sufficient for the binary operation on cosets of $N$ to be closed.
It is in fact necessary too, for if $(gN)(hN) = kN$, we can pick the identity in the first $N$ of $gNhN$ to get $ghN = kN$. Then we can put $h = g^{-1}$ to get $(gNg^{-1})N = N$, which in turn implies $gNg^{-1} \subseteq N$.
A: A left coset $gN$ of a (normal) subgroup $N$ of a group $G$ is a set of elements of $G$, namely, the set
$$gN = \{ gx \mid x\in N\},$$
consisting of all those products of the form $gx$, as $x$ varies over $N$.  This is a subset of the group $G$.  Sometimes, different choices for the element $g$ yield the same coset; that is, $g_{1}N = g_{2}N$, while $g_1\neq g_2$.  Of course, it can also be the case that $g_{1}N \neq g_{2}N$, depending on the choices of $g_1$ and $g_2$.  Therefore, it you collectively take all possible distinct cosets $gN$, for all possible choices of $g\in G$, you get a collection of these subsets of $G$.  This set is called $G/N$.  In other words,
$$G/N = \{ gN \mid g\in G\}.$$
It turns out that the set of distinct cosets form a partition of $G$.  And, the proof of Lagrange's Theorem tells you that the cosets all have the same size, equal to the size of $N$ itself.  The number of these cosets is the index $[G:N]$.  So far, none of this requires that $N$ be normal in $G$.  To turn this collection of subsets of $G$ into a group, it is necessary to define an operation on any pair of them.  So, suppose you have two cosets $aN$ and $bN$ of $N$.  We need to come up with another coset that we're going to call the product of $aN$ and $bN$.  It's pretty natural to just define the product of $aN$ and $bN$ to be the coset $(ab)N$ (or just $abN$ - the parentheses aren't needed).  This is the coset of $N$ corresponding to the element $ab$ of $G$.  Now the point of taking $N$ to be normal is to ensure that this definition is, in fact, a valid one, because the individual factors $aN$ and $bN$ may be represented in different ways.  That is, there might be distinct $a_1$ and $a_2$ in $G$ such that $a_1N = a_2N$.  Likewise, we might have $b_1N = b_2N$, for distinct $b_1, b_2\in G$.  The fact that $N$ is normal is what allows you to conclude that everything is okay.  (See Addendum below.)  The identity element of the group $G/N$ defined in this way is just the coset $N = 1N$ itself because $1N gN = (1g)N = gN$, for any $gN\in G/N$.  The result is a group whose elements are certain kinds of subsets (namely, the cosets of $N$) of $G$.
Now, the really cool part of all of this is that the normal subgroups give you precisely the kernels of homomorphisms from $G$ into other groups.  Because of the way the product is defined in $G/N$, the mapping $\pi$ from $G$ to $G/N$ that sends each $g\in G$ to the corresponding coset $gN\in G/N$ is a homomorphism: $\pi(ab) = abN = (aN)(bN) = \pi(a)\pi(b)$, for any $a,b\in G$.  Furthermore, the kernel of this $\pi$ is the set of elements $g$ in $G$ such that $\pi(g)$ is the identity of $G/N$.  That means $\pi(g) = N$, which by the definition of $\pi$ means that $gN = N$.  This is equivalent to $g\in N$, so $N$ is the kernel of $\pi$.
Conversely, suppose that $\psi$ is some homomorphism from $G$ to another group $H$, and let $N$ be its kernel. If $g$ is any element of $G$, then, for any $x\in N$, we have$$\psi(g^{-1}xg) = (\psi g)^{-1}(\psi x)(\psi g) = \psi(g)^{-1}\cdot 1_{H}\cdot\psi(g) = 1_{H},$$
because $x\in N=\operatorname{ker}(\psi)$ means that $\psi(x) = 1_H$.  Therefore, $g^{-1}xg\in N$ and, since $x$ was arbitrary, it follows that $g^{-1}Ng\subseteq N$.  The reverse inclusion is similar, so in fact $N = g^{-1}Ng$, for any $g\in G$.
Addendum
We want to define the product $aN\cdot bN = abN$ of two cosets $aN$ and $bN$ in $G/N$.  For this to make sense, we need $N$ to be normal.  That is, we need to ensure that, if $a_1, a_2, b_1, b_2\in G$ with $a_1N = a_2N$ and $b_1N = b_2N$, then $a_1b_1N = a_2b_2N$.  (We say that the operation is "well-defined", in this case, because it doesn't matter which group element we chose to form the coset, as long it yields the same coset.)
To show this, remember that two cosets $xN$ and $yN$ are equal, $xN = yN$ if, and only if, $x^{-1}y\in N$ (even if $N$ is not normal).  Now let's apply this.  Since $a_1N = a_2N$, we have $a_{1}^{-1}a_{2}\in N$.  This means that $a_2 = a_{1}u$, for some $u\in N$.  And, since $b_1N = b_2N$, we have $b_{1}^{-1}b_{2}\in N$.  This means that $b_2 = b_{1}v$, for some $v\in N$. To show that $a_1b_1N = a_2b_2N$, we must therefore show that $(a_1b_1)^{-1}(a_2b_2)\in N$.  Let's calculate:
$$(a_1b_1)^{-1}(a_2b_2)
= b_{1}^{-1}a_{1}^{-1}a_2b_2
= b_{1}^{-1}a_{1}^{-1}a_{1}ub_{1}v
= b_{1}^{-1}ub_{1}v.$$
Now we use the fact that $N$ is normal to conclude that $b_{1}^{-1}ub_{1}\in N$, since $u\in N$.  Since, also, $v\in N$, their product $(b_{1}^{-1}ub_{1})v$, which is the last expression in the calculation above, must belong to $N$, as desired.
