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I'm having trouble verifying what ought to be a relatively simple detail in a proof of Milnor's book on algebraic K-theory, in the section on universal central extensions.

Here is the set up. Let $G$ be a perfect group ($G = [G,G]$), and let $\gamma:F \to G$ be a homomorphism from a free group $F$ onto $G$. Let $R \subset F$ be the kernel. Then $[R,F] \subset F$ is a normal subgroup, and we have a surjection $$ \phi:F/[R,F] \to F/R \cong G $$ The kernel of $\phi$ is central, so by a previous lemma of Milnor's, the commutator subgroup $$ (F/[R,F])' = [F,F]/[R,F] $$ is a perfect central extension of $G$. Now we want to show that $[F,F]/[R,F]$ is the universal central extension. Let $(X,\psi)$ be a central extension of $G$. Then by the universal property of free groups, there is a homomorphism $h:F \to X$ over $G$, that is, $\psi \circ h = \gamma$.

All this I can follow. Next, Milnor assers that $h[R,F] = 1$, which I don't see. I know that $\psi \circ h[R,F] = \gamma[R,F] = 1$, so $h[R,F] \subset \ker \psi$. We know that $\psi$ is a central extension, so $\ker \psi$ is contained in the center of $X$. How do I get from here to see that $h[R,F] = 1$?

Joshua Ruiter
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1 Answers1

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We have a surjection $\psi:X\to G$ with kernel $K$, contained in the centre of $X$.

We also have a map $h:F\to X$ with $\phi\circ h=\gamma$. For $r\in R$, then $1=\gamma(r)=\psi(h(r))$, so that $h(r)\in K$.

Let $f\in F$ and $r\in R$ then $h(f)\in X$ and $h(r)\in K$. As $h(r)$ is in the centre of $X$ then $h(f)$ and $h(r)$ commute. Therefore $$h([f,r])=[h(f),h(r)]=1$$ and consequently, $h([F,R])=\{1\}$.

Angina Seng
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