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I'm having a lot of trouble wrapping my head around this one. The mathematical definition is that $G'$ is a factor group of $G$ if $G'$ = $G$/$N$, where $N$ is some normal subgroup of $G$. $R'$ is a factor ring of $R$ if $R'$ = $R$/$I$, where $I$ is an ideal.

My question is, just "what the hell" are they exactly? Are we taking all elements of $N$ (and similarly $I$) and taking them out of $G$ and $R$? If so, why? Why is that important? If not, what exactly are we doing when we create a factor group/ring? I could really use some help visualizing these. Thank you.

Max
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  • Have you studied modular arithmetic? – Jack M Dec 07 '16 at 23:48
  • See http://math.stackexchange.com/a/1817020/589. – lhf Dec 08 '16 at 00:51
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    0 votes, but 5 answers. Strange! Why answer something you don't think is useful? Well I won't ever bother reading those answers then. – Kibble Dec 08 '16 at 01:16
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    Interesting thing some of us can answer questions we don't particularly like (or dislike of course)... It's almost as if this is pretty common knowledge among people on this site @Kibble – user347489 Dec 08 '16 at 01:51
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    I really think the key here is this: Start with an equivalence relation on a group $G$. This partitions $G$ into equivalence classes (which, first and foremost, are simply subsets of $G$.) Given that we can multiply elements of $G$ together, the question is: Can we multiply the equivalence classes together? In particular, we'd like to know if we can treat the equivalence classes as elements of a new group, the "quotient of $G$ module the equivalence relation." – PeterJL Dec 08 '16 at 04:01

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Another important way to think about quotient groups: they are precisely the images of group homomorphisms.

This is a consequence of the first isomorphism theorem: if $G$ and $H$ are groups and $\phi : G \to H$ is a group homomorphism, then $G/K \cong \operatorname{im}\phi$, where $K = \operatorname{ker}\phi$.

Conversely, if $N$ is any normal subgroup of $G$, then there is a natural homorphism $\phi : G \to G/N$ given by $\phi(g) = gN$. The image of this homomorphism is $G/N$, and the kernel is $N$.

Similar remarks apply to quotient rings.

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    Indeed. At first it seems like there could be nonzero homomorphisms from your group into lots of other groups, potentially a proper class of groups. But no, this shows us that the (nontrivial) homomorphisms are only set-sized: you can link them to a subset of the powerset of $G$ (the set of normal subgroups.) – rschwieb Dec 08 '16 at 01:55
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    Would the downvoter care to comment? I find this viewpoint very useful and often less cumbersome than working explicitly with cosets. –  Dec 08 '16 at 01:57
  • I didn't down-vote, but I'll note that every group is the image of a group homomorphism, so your leading comment doesn't really shed much light on the notion of quotient group. On the other hand, your appeal to the First Iso. Theorem is important, and probably much closer to the heart of the matter. But still, my guess is that the OP what's some intuition here. – PeterJL Dec 08 '16 at 03:53
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    @PeterJL Sure, every group is trivially the image of the identity map from itself to itself, just as every group is trivially a quotient of itself by ${1}$. This case isn't very interesting no matter how you view it. My point is that in the nontrivial case, it can often be less confusing/messy to consider a quotient group as the image of a homomorphism instead of as a collection of cosets, and that it is always possible to do so. –  Dec 08 '16 at 05:38
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    @PeterJL For a concrete example, consider the statement "any quotient of a nilpotent group is nilpotent," where we use the definition of a nilpotent group as one which has a central series. This is rather ugly to prove by working directly with quotient groups, but it becomes quite a bit cleaner if we prove the equivalent statement "any homomorphic image of a nilpotent group is nilpotent." –  Dec 08 '16 at 05:51
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Let me give you an example:

$x,y ∈ \mathbb Z \,, \equiv_2$ is an equivalence relation, where $x$ and $y$ have to be both even or uneven ($\bmod2$ ).

Then $\mathbb Z/\equiv_2 = \{[0],[1]\}$ because every Element out of $\mathbb Z$ is in one of those two equivalence classes. $[0] = [2]$ because $0 \bmod 2 = 2 \bmod 2 \cdots$

R_D
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WhoAmI
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If $G$ is a group and $H$ its normal subgroup, then $G/H$ is a group whose elements are of the form $gH,$ i.e. the elements of that group are cosets of $H$. The operation of that group is given by $$gH \cdot g'H=gg'H$$ This operation is well-defined thanks to normality of H.

For rings, quotient rings are defined similarly. The only difference is that we usually use "+" for addition operation and $\cdot$ for multiplication operation. Can you guess how multiplication is defined in a quotient ring?

Pawel
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A primary intuition of quotient groups is "modularity". Suppose we have group $G$ with subgroup $H$. Let us define an equivalence relation on $G$; say $g \sim g'$ if $g + (-g') \in H$. Thus, we consider the set of $\sim$ equivalence classes, $G / {\sim}$. Then (if $H$ is a normal subgroup) we can define a group $(G /{\sim}, +)$, such that addition between equivalence classes $[g], [g']$ is given by $[g+g']$. This group is precisely the quotient group.

Something for you to ponder; why do you think $H$ must be normal in order for $(G/{\sim}, +)$?

Meow Mix
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  • I have ever seen the adjective "modularity" applied to this. But I suppose it's fair to think of the cosets being cohesive pieces in that way. – rschwieb Dec 08 '16 at 01:50
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It is a good question and I will answer it from the perspective of universal algebra, which includes all of these. Any algebraic structure is a set with an imposed structure on it through various $n$-ary operations. You are most commonly familiar with the binary operations.

However a thing we often want to do is to group objects together. In some cases there are simply "too many" where the majority are distinct by some quality we do not really care for. So we try to group these together such that each of these classes contain all elements that are the same with respect to our desired property and differ only by the property we do not care about.

Such classes are equivalence classes and the relation between the elements is an equivalence relation. A prototypical is the rational numbers where we have $$\frac{p}{q}=\frac{np}{nq}$$ In a purely set theoretical view those are distinct and different but we don't care about that in our usage of rationals so they are grouped together. However an equivalence relation does NOT necciserily go together with our $n$-ary operations. This is where the concept of well-definedness comes from. A function is well-defined if the choice within the equivalence class does not matter for the result of the function. An example of a non-well-defined is $$f(\frac{p}{q})=p$$

Now what does this have to do with this? A relation that respects the operations and thereby makes the $n$-ary operations well-defined with the equivalence classes, is called a congruence class. With a congruence relation we can split up the structure into the classes and have the operations be meaningful and well-defined. A fundamental property is that any homomorphism, which is defined for universal algebra as $$\varphi(f^A(x_1,\ldots,x_n))=f^B(\varphi(x_1),\ldots,\varphi(x_n))$$ where $\varphi:A\to B$ is a homomorphism and $f^A,f^B$ are $n$-ary operation in the source adn target structure, is that the kernel $$\ker\varphi=\{(x,y)\in A^2:\varphi(x)=\varphi(y)\}$$ is always a congruence relation. More importantly the CONVERSE is true, every congruence relation IS the kernel of a homomorphism. We can then show that $\varphi(A)=A/\ker \varphi$ for any algebraic structure. Where the quotient is the grouping according to the congruence relation.

For groups and rings it can be shown rather easily that a congruence relation neccesitate a normal subgroup and an ideal and the definitions we use.

As for a visual, imagine the structure as a bread and then you cut it into slices that you focus on, rather than the individual atoms in the bread.

Zelos Malum
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The basic idea is that whenever you have a sub-group(ring), this sub-object allows you to classify the elements of the big group(ring) into different set-theoretical classes or "boxes".

How? In the easiest way you can imagine: consider a sub-group $H\subset G$ and $a,b\in G$. We say these elements are in the same box (class) iff the sets $aH=bH=\{bh:h\in H\}$. The same is done with rings, but using the commutative structure (if $I\subset R$ and $a,b\in R$, then $a\sim b$ iff $a-b\in I$).

Now, these are just set-theoretical decompositions of your original algebraic objects, but if additionally the sub-group(ring) is normal(ideal) these boxes inherit the algebraic structure from the original group(ring)!

user347489
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