Normal groups: Abstract question about abstract algebra. Let $G$ be a group, and $H\triangleleft G$ be a normal subgroup of $G$.
To my best understanding,the group $^G/_H$ is often used to "simplify" groups, meaning it enables us to deal with a simple form of $G$, so in other words, throwing all the elements in $H$ together and labeling them as $e_{G/H}$ (the unity of $^G/_H$), is to say that $H$ is, in some sense, "bothering us", or "interfering" when trying to investigate $G$.
In what way is $H$ "bad"?
I feel like there's something deep that I'm missing here...
 A: Groups arise very naturally as symmetries of geometric objects, for example as in the comment of @MarcinLos. 
More generally, groups arise as actions: a group $G$ can "act" on some geometric object. The difference between "symmetry groups" and "actions" is that when a group $G$ acts on a geometric object $X$, some group elements might act trivially on $X$, meaning that they act on $X$ as the identity map. The subset of $G$ which acts as the identity map on $X$ is a normal subgroup called the "kernel" of the action. The quotient group $G/$kernel can then be thought of as a true symmetry group, in which each group element acts nontrivially. So in this case the "bad" elements of $G$ are the elements of the kernel, the elements that are acting trivially on $X$.
There are zillions of examples of this phenomenon. Here is just one example, one of my favorites: the group $SL_2(\mathbb{R})$, which is the group of $2 \times 2$ matrices with real coefficients and determinant $1$. This group acts on $\mathbb{R}^* = \mathbb{R} \cup \{\infty\}$ by "fractional linear transformations":
$$\begin{pmatrix}a&b\\c&d\end{pmatrix} r = \frac{ar+b}{cr+d}
$$
(and if you ever learn about hyperbolic geometry $\mathbb{H}$ modelled as the upper half of the complex plane, one can turn this into an action on $\mathbb{H}$ by replacing the $\mathbb{R}^*$ variable $r$ with a complex variable $x+iy$, $y >0$).
Of course the identity matrix $I$ acts as the identity map on $\mathbb{R}^*$. More "bothersome" is that the negative of the identity matrix
$$-I = \begin{pmatrix}-1&0\\0&-1\end{pmatrix}
$$ 
also acts as the identity map on $\mathbb{R}^*$. Fortunately, without too much trouble one can prove that $\pm I$ is the kernel of the action: these are the only matrices which act as the identity map on $\mathbb{R}^*$. 
So if you are bothered by $\{\pm I\}$ acting as the identity, you may want to "simplify" the action by taking the quotient group:
$$PSL(2,\mathbb{R}) = SL(2,\mathbb{R}) / \{\pm I\}
$$
And now you get a true symmetry group, each non-identity element of which acts on $\mathbb{R}^*$ (and on $\mathbb{H}$) as a non-identity map.
A: $G/H$ is not exactly a simpler form of $G$, and it has (in general) properties different from those of $G$. Consider what are probably the simplest examples of quotient groups - $\mathbb{Z}_n = \mathbb{Z}/n\mathbb{Z}$. The subgroup in question, $n\mathbb{Z}$, is not "bad" in anyway. But if we are considering all integers mod $n$, there is no point looking at, say, $n + 1$ and $2n + 1$ as distinct elements.
For example, in $\mathbb{Z}_3$ consists of the three cosets (or equivalence classes) $\bar{0} = \lbrace\ldots -3, 0, 3, 6, \ldots\rbrace$, $\bar{1} = \lbrace\ldots -2, 0, 4, 7, \ldots\rbrace$, and $\bar{2} = \lbrace\ldots -1, 0, 5, 8, \ldots\rbrace$. Now, $4$ and $7$ are, for all purposes, identical, because $x + 4$ is in the same equivalence class as $x + 7$ (which is the same as the class of $x + 1$, for that matter). Then it is only a matter of convenience and simplicity to think of $\mathbb{Z}_3$ as $\lbrace \bar{0}, \bar{1}, \bar{2}\rbrace$. The elements $0$, $1$, and $2$ represent their equivalence classes in every manner desirable.
To take a real-world example, you're doing modular arithmetic (and therefore using quotient groups) when telling time. You might think of the timeline as a number line (i.e., as the group $\mathbb{Z}$) with, say, the beginning of $0$ AD as the point $0$. Then the set of midnights ($0$ hour) forms a (normal) subgroup. When we say "it's midnight", we don't mean it's midnight of the first day of $0$ AD. We mean (roughly) "some multiple of 24 hours have elapsed since the midnight of $0$ AD". But see how cumbersome and unnecessary that is! It is much better to divide a day into twenty four hours and make the clock read $00:00$ at the twenty fourth hour.
Apart from this, studying the structure of quotient groups tells us some things about the group. They are also important when studying homomorphisms, where we construct quotient groups using the kernel of the homomorphisms.
