# Upper central series, semidirect products and a lifting property (argument verification)

Let $G$ be a profinite group, and define the lifting property: $(*_p)$ For every extension $1 \to P \to E \to W \to 1$ where $E$ is finite and $P$ is a $p$-group and for every surjective morphism $f\colon G \to W$ there is a surjective extension $f'\colon G \to E$.

Suppose $G$ satisfies $(*_p)$ in the following two cases:

• (i) Every subgroup of $E$ which projects onto $W$ is equal to $E$.
• (ii) $P$ is abelian and killed by $p$, and $E$ is isomorphic to the semidirect product $P \rtimes W$

Then, I want to prove that $G$ satisfies the general case. My progress so far was taking the long upper central series $P_i$ for $P$ (with $P_0 = 1$ and $P_n = P$ for some $n \in \mathbb{N}$), and noticing that we have surjections $$E = E/P_0 \to E/P_1 \to \cdots \to E/P_{n-1} \to E/P_n = E/P = W \to 1$$ with kernels $P_i/P_{i-1}$ that are abelian groups killed by $p$. If we have the lifting property for the extension $$1 \to P_n/P_{n-1} \to E/P_{n-1} \to W \to 1$$ then we have it for the general case. I'm not sure how can I use (i) and (ii) now.. If $p \not\mid |W|$ than (ii) easily applies, but that need not be the case.

For context, this is part of exercise 1 in section 3.4 of chapter 1 in Serre's Galois Cohomology.

PROGRESS I know how to reduce the problem a little using (i). Using the reductions as before, we can suppose that $P$ is abelian and killed by $p$. It is a known fact that $P$ has a minimal subgroup complement $K$ in $E$: $PK = E$ and for any $L \leq K$ s.t. $PL = E$ we have $L = K$. But this means that the extension $$1 \to P \cap K \to K \to W \to 1$$ satisfies (i), and thus we have a surjective extension of $f\colon G \to W$ to $\tilde{f}\colon G \to K$. How can we use (ii) now to lift it to $E$?

FINAL STRETCH We can suppose (WLOG by the remarks above, I guess) that $P = \mathbb{Z}/p\mathbb{Z}$. We have the minimal complement $K$ and a lifting $\tilde{f}\colon G \to K$ of $f\colon G \to W$. If $K \neq E$ then the intersection $K \cap P$ cannot equal $P$ and we must therefore have $K \cap P = 1$, proving that there is an isomorphism $E = P \rtimes K$. Then, we apply (ii) to get a lifting $f'\colon G \to E$. I am just unsure about reducing to $\mathbb{Z}/p\mathbb{Z}$...

Actually, maybe the reduction isn't even necessary... if $K \neq E$ then $K \cap P$ must be a proper subgroup of $P$. But it is a normal subgroup of $E$, and by the minimality of $P$ it must be trivial!