# Why do normal subgroups “work”? [duplicate]

This is something I've wondered about for a while now; What is so special about normal subgroups that makes modding out by them "act nice"?

I understand the proofs for things like the first isomorphism theorem, but all these proofs really seem to do is verify that we can mod out by any sub-group satisfying $$gNg^{-1} = N$$ and end up with another group. I get the verification; that's not what I'm asking about.

It's possible to mod out by any subgroup, but what's so special about the property $$gNg^{-1} = N$$ that makes it so that we can "move" $$N$$ around inside $$G$$ in a way that satisfies the group axioms? Is there something deeper about the normality property that makes this make sense?

• So suppose you have a subgroup $H \le G$. You can form the set of $H$-cosets $\{gH : g \in G\}$. You want to turn it into a group, and there is an obvious choice for the group operation, namely $(gH)(g'H) := (g g') H$, but you have to check that it is well defined. This holds precisely if $H$ is normal. – Nate Eldredge Jan 10 at 0:00
• Yeah. That part I totally understand. I'm just wondering what's so special about normality that makes the operation well defined. I might be asking a question that doesn't have any kind of deeper answer; Perhaps it's just a brute fact. But when I think of the group $G/N$ as "the group consisting of ways to move $N$ inside of $G$", I'm left puzzled why $N$ being normal gives this collection of movements a group structure, but movements of a non-normal subgroup don't have a group structure. – Bears Jan 10 at 0:07
• If you multiply $H$ with itself, you get $H$ again (because $H$ is a subgroup). Therefore to get $gHg'H = gg'H$, you just have to move $g'$ past the left $H$. But that being allowed (that is, $g'H=Hg'$) is exactly the definition of a normal subgroup. – celtschk Jan 10 at 0:18
• Normal subgroup happen to “code” congruences, which are the real thing you are moding out by; it just so happens that ideals and normal subgroups “work” for groups and rings. If you work with other types of structures, such as semigroups or monoids, you actually do need to work with the equivalence relation itself. See this extensive answer about it – Arturo Magidin Jan 10 at 4:05
• @BenBlum-Smith: As you can see, I waited before voting to close (knowing my vote was binding). Bears has been active several times since my comments, and did not speak up, so I went ahead. But just like I can close with a single vote, I can also re-open with a single vote if I’m not mistaken. If he speaks up, I’ll be happy to re-open. – Arturo Magidin Jan 10 at 16:50

A general quotient set is done by an equivalence relation $$\sim$$ on a set $$S$$, and we can informally think about $$S/\sim$$ as the set obtained by keeping the elements but replacing the original equation of $$S$$ by $$\sim$$, so that $$x\sim y$$ in $$S$$ means $$x=y$$ in $$S/\sim$$.

Now, indeed any subgroup $$H\le G$$ determines an equivalence relation by taking the partition of left (or right) cosets, so nothing stops us to define the quotient set $$G/H$$ as $$G/\sim_H$$ where $$x\sim_H y\iff y^{-1}x\in H$$.

However, in order for it to inherit the group structure, we also need the induced equivalence relation to respect the group operation, and in particular we need $$gg^{-1}=1$$ to hold.
But in $$G/H$$ we will have $$h=1$$ for any $$h\in H$$, so we must have $$ghg^{-1}=g1g^{-1}=gg^{-1}=1$$ in $$G/H$$, which really means $$ghg^{-1}\sim_H 1$$ in $$G$$, i.e. $$ghg^{-1}\in H$$.

It turns out that this additional condition (of being normal) on $$H$$ is sufficient as well: in that case the group structure is inherited to the quotient set.

A more formal way is that normal subgroups are exactly the kernels of some homomorphism (where kernel is the preimage of $$1$$).

• It’s really that normal subgroups “code” congruences; the kernel notion doesn’t work for monoids, because you need to capture the entire equivalence relation $x\sim y\iff f(x)=f(y)$; with groups, modules, and rings, you can code it with a subset of the original structure, but for more general structures you can’t. So kernels work only because you have a lemma that says that $f(x)=f(y)$ if and only if $xy^{-1}$ is in the kernel. – Arturo Magidin Jan 10 at 4:21
• That's correct, but for general algebraic structures, the kernel of a homomorphism $f$ is defined as the equivalence relation $x\sim y\iff f(x)=f(y)$. @arturomagidin – Berci Jan 10 at 9:01

Concerning your comment about "the group consisting of ways to move N inside of G", the group structure is actually being given to a way of moving the cosets of $$N$$ around rather than $$N$$ itself.

A standard way of moving subgroups themselves around is by conjugation:- $$H$$ going to $$g^{-1}Hg$$.

Here one can see something very special about normal subgroups. They are precisely the subgroups which do not move around at all!

Furthermore this is the property used in standard proofs that the group operation on cosets is well defined.