# A projectivity property in the category of groups

Let $$F_r$$ be the free group on $$r$$ generators, let $$G$$ and $$H$$ be finite groups, and let $$F_r\xrightarrow{\alpha}H\xleftarrow{\beta}G$$ be surjective homomorphisms. It is then easy to see that we can choose $$\phi\colon F_r\to G$$ with $$\beta\phi=\alpha$$, but $$\phi$$ need not be surjective, especially if $$G$$ is too big. Now suppose in addition that there exists a surjective homomorphism $$F_r\to G$$. Is it then possible to find a surjective homomorphism $$\phi\colon F_r\to G$$ with $$\beta\phi=\alpha$$?

I think that this works if $$G$$ is nilpotent, because we can then reduce to the case of $$p$$-groups, and use the Frattini quotient to detect surjectivity. However, it would be tidier if I could prove it without any hypothesis. It also works when $$r=1$$, but even that case requires some modular arithmetic and so is not completely trivial. On the other hand, I think that in practice a randomly chosen $$\phi$$ will be surjective with high probability.

• "It also works when $r=1$": but then $G$ is abelian and hence nilpotent, so isn't it a particular case of the preceding assertion?
– YCor
Commented Aug 8 at 11:19
• @YCor Yes, you can think of $r=1$ as a special case of the nilpotent case, or you can do it directly which is only a tiny bit different. Commented Aug 8 at 11:21
• I think yes and you should look at embedding problems and Iwasawa's criterion for freeness for free profinite groups and use the free group is dense. I don't have time to answer now or check that my memory is correct. Commented Aug 8 at 11:49
• Perhaps it's worth noting that this is false without the assumption that $G$ and $H$ are both finite. Take any group with two distinct Nielsen classes of generating sets. For instance, the dihedral group $D_{10}$ has two distinct classes of generating sets of cardinality 2 (en.wikipedia.org/wiki/Nielsen_transformation). Take $\alpha$ and $\beta$ to be the corresponding maps $F_2\to D_{10}$.
– HJRW
Commented Aug 8 at 11:53
• @HJRW or more simply the two maps $\mathbf{Z}=F_1\to \mathbf{Z}/5\mathbf{Z}$, $1\mapsto 1$, $1\mapsto 2$.
– YCor
Commented Aug 8 at 14:55

Clean Answer. This is known as Gaschutz's Lemma and was proved in W. Gaschutz, Zu einem von B.H. und H. Neumann gestellten Problem, Math. Nachr. 14 (1956), 249-252.

It says if $$G$$ and $$H$$ are finite groups, $$f\colon G\to H$$ is an epimorphism and $$h_1,\ldots, h_n$$ are generators of $$H$$ where $$n\geq d(G)$$ with $$d(G)$$ the minimal number of generators of $$G$$, then you can find lifts $$g_1,\ldots, g_n$$ of $$h_1,\ldots, h_n$$ which generate $$G$$.

The proof uses an inclusion-exclusion inductive counting argument.

What you want then follows immediately by lifting the images of the generators of $$F_r$$.

Original Answer. The answer is yes. Look at Theorem 3.5.8 in Ribes and Zalesskii's book Profinite Groups (2nd edition). If we restrict to $$\mathcal C$$ the formation of all finite groups, it says that a profinite group $$F$$ is a free profinite gruop of rank $$m$$ if and only whenever you have a surjection $$A\twoheadrightarrow B$$ of finite groups with $$A$$ of rank at most $$m$$ and a continuous surjection $$F\to B$$, there exists a continuous surjective lift $$F\to A$$.

Since a free group embeds densely in the free profinite group of the same rank and since finite groups are discrete, the result you want follows.

The key point is Proposition 2.5.4 of Ribes and Zalesskii, which is I believe due to Gaschutz. They prove that if $$G$$ and $$H$$ are finite groups, $$f\colon G\to H$$ is an epimorphism and $$h_1,\ldots, h_n$$ are generators of $$H$$ where $$n\geq d(G)$$ with $$d(G)$$ the minimal number of generators of $$G$$, then you can find lifts $$g_1,\ldots, g_n$$ in $$G$$ which generate $$G$$. This uses an inclusion-exclusion inductive counting argument.

• Thanks. Now I just need to work out who has borrowed our library copy of that book! Commented Aug 8 at 12:41
• You can probably find this in books of finite group theory under the name Gaschutz Lemma. I had forgotten that this was the crux of Iwasawa's criterion. Commented Aug 8 at 12:44
• Thanks, I have found the Gaschutz paper now, and also someone has sent me a PDF of Ribes and Zaleskii. Commented Aug 8 at 13:06
• You are welcome. Commented Aug 8 at 13:21