Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It goes like this:

Let $G$ be a group, $H < G$, and $N \lhd G$ be the smallest normal subgroup containing $H$. Then any $f \in \operatorname{Hom}(G,G')$ such that $H$ is in its kernel uniquely factors through $G/N$.

It's a simple proposition, but it's so weirdly formulated that I spent the whole day picking it apart.

Basically, it says two things:

1) $H < N \lhd G$, and $N < \operatorname{Ker}f$ (so $N \lhd \operatorname{Ker}f$), which is trivial to prove.

2) Any hom uniquely factors through a quotient by any normal group that is a subgroup of its kernel.

So why combine two natural propositions into a single one that is so weird? Is it used in some important place or does it have any significance in the group theory by itself? Or am I mistaken and the second proposition I put forward is generally false?

share|improve this question
1  
The correct spelling is "weird." –  Qiaochu Yuan Feb 22 '11 at 10:39
    
Thank you for the correction :) –  Alexei Averchenko Feb 22 '11 at 10:51
    
What's $G^\prime$? –  Peter Taylor Feb 22 '11 at 12:43
    
@Peter : here it's just some other group. –  Alexei Averchenko Feb 22 '11 at 13:57
    
In your list of things that the result "basically says" you are forgetting one thing: $N$ is the smallest normal subgroup of $G$ that contains $H$. –  Arturo Magidin Feb 22 '11 at 14:12
add comment

2 Answers

up vote 9 down vote accepted

I suspect you find Lang's statement weird because you are thinking of it as a technique for generating homomorphisms, and then wondering why Lang is putting a strange emphasis on these particular kinds of normal subgroups.

But to appreciate the proposition, you should think from a different point of view. E.g. suppose that you are given a homomorphism $f: G \to G'$, and that you are told that $f(h) = 1$ for every $h \in H$. Then what can you conclude about $ker(f)$? Well, it certainly contains $H$, since this is what you are told. But since the kernel is normal, it in fact contains the normal subgroup of $G$ generated by $H$, which is what Lang calls $N$.

As a side remark: although $N$ is correctly described as the smallest normal subgroup containing $H$, it is probably better (from a psychological point of view) to think of $N$ as the normal subgroup of $G$ generated by $H$; this helps bring out the role of $H$.

share|improve this answer
1  
To help support your last argument, $N$ is usually called the normal closure of $H$: en.wikipedia.org/wiki/Normal_closure –  lhf Feb 22 '11 at 13:11
add comment

Well, I don't know Lang's reason, so I can only guess. First of all, it is a concise and easy enough statement and I don't see what's so particularly weird about it.

That said, I think the main reason is that the proposition as stated by Lang accurately describes the categorical image $\langle\langle H \rangle\rangle_{G}$ (the normal closure of $H$ in $G$, or as your proposition states the kernel of the cokernel) of the inclusion $H \hookrightarrow G$ and its categorical cokernel $G/\langle\langle H \rangle\rangle_{G}$ when the inclusion is viewed as a morphism in the category of groups, see e.g. here.

As for your last question, your second proposition is correct and follows from the usual homomorphism theorem.

Added much later: It seems that $\langle H^{G} \rangle$ is the usual notation for the normal closure which I denoted $\langle\langle H \rangle\rangle_{G}$ — some people insisted on that by editing it in — but in my opinion this is really ugly and, much worse, quite ambiguous.

share|improve this answer
    
I thought it was weird because it starts with $H$ and then never really uses it; it is also kind of deceptive, it hints that you have to pick a special kind of normal subgroup, but it's not the case (as can be demonstrated by taking a normal $H$). –  Alexei Averchenko Feb 22 '11 at 11:16
    
@Alexei: Well, it uses and/or includes the fact that there is such a thing as a smallest normal subgroup $N_{G}(H)$ of $G$ containing $H$. –  t.b. Feb 22 '11 at 11:20
    
I can't see it: it is really just a reformulation of the homomorphism theorem you mentioned (thanks, BTW :)) that pretends to be its special case. –  Alexei Averchenko Feb 22 '11 at 11:41
2  
Theo: $N_G(H)$ generally denotes the normalizer in $G$ of $H$, which could be described as the largest subgroup of $G$ in which $H$ is normal. What we are talking about here is the normal closure of $H$ in $G$, which is usually denoted by $\langle H^G \rangle$. –  Derek Holt Feb 22 '11 at 12:33
1  
Urgh. That notation clashes fixed points :/ –  Mariano Suárez-Alvarez May 28 '11 at 23:12
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.