# No lifts in an exact sequence of profinite groups?

In pg. 24 of his book on Galois cohomology, Serre gives the following exercise:

"Give an example of an extension $1 \to P \to E \to G \to 1$ of profinite groups with the following properties:
(i) $P$ is a pro p-group
(ii) $G$ is finite.
(iii) A Sylow p-subgroup of $G$ lifts to $E.$
(iv) $G$ does not lift to $E.$"

I am not sure how this can happen, allow me to explain. Let us start by assuming that all groups involved are finite. Suppose that $S$ is a Sylow $p$-subgroup of $G$ and that it lifts to $E.$ Taking the fiber product along the inclusion $S \to G,$ we get the exact sequence $$1 \rightarrow P \rightarrow E \times_G S \to S \rightarrow 1.$$ By the assumption that $S$ lifts to $E,$ this exact sequence is split.
If $P$ is an abelian $p$-group, this should imply that there is a lift of $G$ to $E,$ since the map $res:H^2(G,P) \to H^2(S,P)$ is injective by the assumption that $S$ is a Sylow $p$-subgroup.

Now, it seems to me that using the above argument for when $P$ is an abelian $p$-group, we should be able to do some devissage argument to show that given any extension $1 \to P \to E \to G \to 1$ of profinite groups and a lift of a Sylow $p$-subgroup of $G$ to $E$ as in the exercise, there is a lift of $G$ to $E$ as well. But this is in contradiction to the exercise.

Obviously, this means I have made a mistake in my thinking, but I have spent some time on it but can not seem to find it. Could someone help me finding where this mistake is?

Update I now see where my mistake lies. I will post this as an answer tomorrow.

• Also, at least in my copy of Serre, an example already appears afterwards in brackets ($G = \mathrm{SL}_2(\mathbb F_p)$ and $E = \mathrm{SL}_2(\mathbb Z_p[w])$, where $w$ is a primitive $p$th root of unity, when $p > 5$); so probably the first thing to do is to see how your (conjectural?) argument treats that example. – LSpice Feb 14 '18 at 21:01
• p22 in the original LNM French version – YCor Feb 14 '18 at 23:02

## 1 Answer

Here is the problem with my conjectural argument.

My hope that one could reduce this to the case where $P$ is finite abelian, but this does not work. This reduction can however be made if one further assumes that $H^1(G,P) \to H^1(S,P)$ is an isomorphism for any abelian $p$-group. To be able to reduce the general case, it turns out that one need to be able to extend the given embedding $s: S \rightarrow E$ to an embedding $G \to E.$ This can be done under the given assumptions and then the devissage argument works.

• If you're satisfied with the answer in mathoverflow.net/questions/293045/…, could you edit your answer with a link to it, and accept your own answer (so that the question is considered as answered and does not periodically reappear in the main list)? – YCor Feb 15 '18 at 18:44