What's the intuition on normal subgroups/ideals 'factoring' out properties in quotient groups/rings? I think 'factoring' out here may be a bit of a misnomer; I mean the qualitative 'removal' of certain unwanted properties. For example, consider the Abelian group $G$, and the subgroup $H$ that is all elements of finite order in $G$. Then $G/H$ has no element (other than identity) of finite order. In some sense selecting this particular subgroup $H$, which consisted of elements of finite order, leads to a quotient group of elements having infinite order.
That was the example given in my book, and I followed it fine and it makes sense in this particular example. There were a couple of others which also made sense, in each particular case. However I cannot quite grasp the intuition behind this in general. I don't see why a normal subgroup or ideal having a particular property tends to mean the quotient group/ring will not.
My book implies this is of great practical importance for the construction of homomorphic images with properties we want, so I am trying to figure out why this is the case? I can see in individual cases, when given proof, why the results hold. But I have zero intuition for why any general property can effectively be eliminated by carefully choosing the subgroup or ideal to have said property?
 A: When $G$ is nonabelian, its elements of finite order are not a subgroup in general (if $g$ and $h$ have finite order, $gh$ could have infinite order).  When $G$ is abelian and $H$ is its subgroup of elements of finite order, the fact that $G/H$ has no nontrivial elements of finite order requires an argument.  To appreciate this, if a group $G$ has a center $Z$, which is always a normal subgroup, the quotient group $G/Z$ might have a nontrivial center.  Passing from $G$ to $G/Z$ does not have to "kill the center" in $G/Z$ because the center of $G/Z$ is represented by elements of $G$ that commute with everything in $G$ "up to multiplication by elements of $Z$", so when $Z$ is killed off, you can be left with nontrivial elements of $G/Z$ that now commute with everything in $G/Z$.
Example: $G = D_4 = \langle r,s\rangle$ where $r^4 = 1$, $s^2 = 1$, and $srs^{-1} = r^{-1}$.  The center of $G$ is $Z = \{1,r^2\}$ and $G/Z$ is a group of order 4, which is abelian (all groups of order 4 are abelian) and therefore $G/Z$ doesn't have a trivial center: the whole quotient group $G/Z$ is its center.  What is going on here is that when you pass from $G$ to $G/Z$, two elements $g$ and $h$ in $G$ that may not commute in $G$ could commute in $G/Z$ because in $G$ they only commute "up to an element in the center": $hg = zgh$ for some nontrivial $z$ in $Z$. Then $g$ and $h$ do not commute in $G$ but their images in $G/Z$ do commute in $G/Z$.
A: I also had issues understanding this until I saw quotients in topology. The idea of a quotient in topology is that you "glue" together stuff. For instance, if you start with the square $I^2 = [0,1]^2$, and you glue together the boundary, you get something like a sphere. That's because what you did was glue together the boundary to a single point. Now all the square is reduced to a point.
In algebra, while the visual intuition is not as clear, the same thing happens. Maybe, to get some intuition, start with the obvious quotients. What's $G/G$? Well you're "gluing" all of $G$ into a single point. This is why $G/G = \{e\}$. So in a sense, when you have a normal subgroup that has a certain property, when you quotient out by that subgroup, you effectively reduce all elements that satisfy that property to a single point.
To give you another example from ring theory, the set of nilpotent elements in a ring form an ideal, called the nilradical $N(A)$. In other words, $x\in N(A)$ if and only if $x$ is nilpotent. So what happens when you consider $A/N(A)$? Well, effectively you reduce every nilpotent element to a point, which makes your ring reduced. There's only one nilpotent element now, namely, $0$.
If you think of this geometrically, if the set of elements satisfying a property $P$ forms an ideal $I$, then $A/I$ has only one element that represents $I$. This is why the property "disappears" in the quotient, because you've reduced it to a point. Indeed, if $x$ satisfied $P$ before, then $x\in I$, and the elements of $A/I$ are of the form $a+I$. But $x+I$ is just $I$, the identity in $A/I$.
A: 'Removing elements' is not the best intuition for quotients, rather to 'make some of them equal', i.e. informally we can think it as keeping the same elements but modifying the equality rules.
We can quotient out a set by any equivalence relation (reflexivity, symmetry and transitivity are just the main - and characteristic - properties of equality).
A prominent example is the high school definition of vectors on the plane: they are directed segments, but two such are considered equal when they form the opposite sides of a parallel heading towards the same direction.
Another example in every day life is the time. We simply consider 23:00 = 11:00pm. They are equal modulo 12.
If we have some algebraic structure (i.e. group or ring), we want the equality relation to keep the operations as well, so we need an equivalence relation that also preserves the operations, those are called congruence relations.
In a group, a congruence relation is determined by the equivalence class $N$ of the identity element $1$, and this is always a normal subgroup:
In the quotient group, with the above informalism, we will have $1=1$ so $1\in N$;
If $x,y\in N$m i.e. if $x=1$ and $y=1$ then $xy=1$ so $xy\in N$ and for any $g$ if $x=1$ then $gx=g$ and $gxg^{-1}=gg^{-1}=1$ so $gxg^{-1}\in N$.
This can be made precise in many ways, e.g. using $[\cdot]$ brackets or pasting $N$ to denote equivalence classes [=cosets] in $G$ instead of the same elements with a new equality.
Another way is using homomorphisms: if $x\in N$, we don't directly say $x=1$ (or correctly, $[x]=[1]$ or $xN=N$) but that $x$ will be equal to $1$ somewhere, i.e. $f(x)=f(1)=1$ for some homomorphism $f:G\to H$.
Indeed, a subset $N$ of a group $G$ is a normal subgroup if and only if it is the kernel of a homomorphism $f:G\to H$.
[Note that the notion of kernel can be generalized to sets or arbitrary algebraic structures simply by taking the congruence relation $g\sim g'\iff f(g)=f(g')$.]
Rewriting the above, $f(x)=1$ means that $x\in N$,
Clearly we have $f(1)=1$ so $1\in N$; if $f(x)=f(y)=1$ then $f(xy)=1$; and if $f(x)=1$ then $f(gxg^{-1})=f(g)f(x)f(g)^{-1}=f(g)f(g)^{-1}=1$.
The converse of the theorem follows by the quotient construction $G/N$, which defines a group operation on the set of cosets.
The congruence relations of rings are determined by the equivalence class of $0$, and that is always an ideal.
That's why we quotient out by normal subgroups and ideals.
