# If one quotient group factors through another, what can we say about the corresponding subgroups?

I was toying with solvable Lie algebras when I stumbled upon a fact that I can't make sense out of, and I suspect I made an error.

I will be talking about groups here, because this is a category that's more familiar to most people.

We know that abelianization has universal property: let $f: G \to A$ be a morphism from a group $G$ to an abelian group $A$. Then it factors through the abelianization $G' := G / [G, G]$ of the group $G$.: $$f: G \ \xrightarrow{\mathrm{ab}_G} \ G' \ \xrightarrow{f'} \ A.$$ Now let's consider a normal subgroup $H \lhd G$ (with the embedding $i: H \to G$) such that $G/H$ is abelian. Then the projection $p: G \to G/H$ factors through $G'$: $$p: G \ \xrightarrow{\mathrm{ab}_G} \ G' \ \xrightarrow{p'} \ G/H.$$ Then what can we say about relation between $H$ and $G'$? My intuition is that we should reverse arrows and then we get $H \leq [G, G]$, but because $p = p' \circ \mathrm{ab}_G$ we actually have $$[G, G] = \ker \mathrm{ab}_G \leq \ker p = H.$$ Did I make a mistake or is this all correct? If it's the latter, then how to make sense out of it?

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Yes this is correct. Would you be happy with: $G/H$ is abelian if and only if $[G,G] \leq H$, so $G'$ is the largest abelian quotient of $G$? In fact, this is just the homomorphism theorem: If $H,K$ are normal subgroups of $G$ then $G \to G/K$ factors through $G \to G/K$ if and only if $K \leq H$. –  t.b. Jul 7 '11 at 11:05
Yes, it makes sense now (when I think about how existence of epis and monos can be translated into informal 'larger then' and 'smaller then'), but why does the arrow-reversing intuition fail here? –  Alexei Averchenko Jul 7 '11 at 11:20
For example, a group $G$ is said to be perfect if $G=G'$ (where $G'$ is the commutator subgroup of $G$). Exercise: prove that if a group $G$ has a subgroup $H$ of index $2$, then $G$ is not perfect. Is the result of this exercise true if $2$ is replaced by $3$ in the previous sentence? Prove that if $G$ is a finite group, if $p$ is the smallest prime divisor of the order of $G$ and if $G$ has a subgroup of index $p$, then $G$ is not perfect. (Hint: think about the core of a subgroup in a group.) Finally, prove Grun's lemma: the center of $G/\textbf{Z}(G)$ is trivial if $G$ is perfect. –  Amitesh Datta Jul 7 '11 at 11:24
If you have a map from $G/M$ onto $G/N$, and you "reverse the arrows", you get a map from $N$ onto $M$, so you expect $N$ to contain $M$, no? –  Gerry Myerson Jul 7 '11 at 13:13

One definition (the "top down definition") of "commutator subgroup of $G$" is that it is the smallest normal subgroup $N$ of $G$ such that $G/N$ is abelian; that is, it is the intersection of all normal subgroups $M$ such that $G/M$ is abelian. Under that definition, the fact that if $G/H$ is abelian then $H$ must contain the commutator subgroup is immediate.

The other definition of "commutator subgroup of $G$" (the "bottoms-up definition") is as the verbal subgroup corresponding to the word $[x,y]$; that is, the subgroup of $G$ generated by all values of the word $[x,y]=x^{-1}y^{-1}xy$, as $x$ and $y$ range over all possible elements of the group. In particular, this verbal subgroup is contained in the kernel of any map from $G$ into a group that satisfies the word, that is, any group for which $[a,b]=1$ for all $a,b$; this is exactly the abelian groups as well, so again you get the result.

These arguments hold for any normal subgroup with the desired property: pick a family $\mathcal{F}$ of groups which is closed under subgroups, isomorphic copies, and arbitrary direct products (i.e., a quasivariety); if $N_i$ is a family of normal subgroups of $G$ such that $G/N_i$ is in $\mathcal{F}$ for each $i$, then so is $N=\cap N_i$; so $N$ is normal in $G$ and has the "universal property relative to $\mathcal{F}$": for any homomorphism $f\colon G\to F$ with $F\in\mathcal{F}$, we have that $N\lt \mathrm{ker}(f)$.

Quasivarieties of groups are also defined via quasi-identities: conditions of the form saying that if a certain finite family $s_i$ of words are trivial, then so is a particular word $s$; i.e., statements of the form $$s_1=1\land s_2=1\land\cdots\land s_n=1\Longrightarrow s=1.$$ They have a corresponding "quasiverbal subgroups", which becomes the verbal subgroup when $n=0$ (and we are dealing with a variety).

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Alexei, for the general case, if $N \lhd G$, then $(G/N)'$ $=$ $G'N/N$ being isomorphic to $G'/(N \cap G')$.

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