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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?

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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
up vote 10 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 $H$.

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To help support your last argument, $N$ is usually called the normal closure of $H$: – lhf Feb 22 '11 at 13:11

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.

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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
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
Urgh. That notation clashes fixed points :/ – Mariano Suárez-Alvarez May 28 '11 at 23:12

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