$U^s = (U^p)^s$ and there exists a normal series with factor group a p-group Let $U$ a finite group.
We can define $U^p$ the smallest normal subgroup of $U$ st $U/U^p$ is a $p$-group and analogously $U^s$ the smallest normal subgroup of $U$ st $U/U^s$ is solvable (both thanks to Arturo Magidin)
$$U^s:=\bigcap\{H\trianglelefteq U \mid U/H \text{ is solvable} \}$$
$$U^p:=\bigcap\{H\trianglelefteq U \mid U/H \text{ is a }p{\rm -group} \}$$
I would like the following to be true

$$U^s = (U^p)^s$$

i think that $(U^p)^s \trianglelefteq U $, then if $U/(U^p)^s$ were solvable then $U^s \subseteq (U^p)^s$
On the other hand

There exists $U_i \leq U,$ $i\in \mathbb{I}_n$ st $$U=U_0 \trianglerighteq U_i \trianglerighteq \cdots \trianglerighteq U_n = U^s,$$ and $U_{i-1}/U_i$ is a $p_i$-group

I know that the derived series has finite length. Then
There exists a minimum $n\in \mathbb{N}$ such that $U^{(n+1)}=U^{(n)}$ and
$$U=U^{(0)} \trianglerighteq \cdots \trianglerighteq U^{(n)}$$
I expected that the subgroup $U^{(n)}$ would have $U^s$ as a subgroup, but what confuses me is that the factor group must be a $p_i$ group.
Any hint is welcome
 A: The equality holds.
Since every finite $p$-group is nilpotent, and nilpotent groups are solvable, it follows that $G/U^p$ is solvable, so $U^s\leq U^p$.
Now, $U^p/U^s$ is a subgroup of $G/U^s$, which is solvable. Since subgroups of solvable groups are solvable, it follows that $U^p/U^s$ is solvable; therefore, $(U^p)^s\leq U^s$.
Since $U^s/(U^p)^s$ is a subgroup of $U^p/(U^p)^s$, it follows that $U^s/(U^p)^s$ is solvable.
Note that $(U^p)^s$ is in fact characteristic in $U^p$: if $\phi$ is an automorphism of $U^p$, and $N\triangleleft U^p$ is such that $U^p/N$ is a solvable group, then $\phi$ induces an isomorphism $U^p/N\cong \phi(U^p)/\phi(N) = U^p/\phi(N)$, so $\phi(N)$ is also a normal subgroup such that $U^p/\phi(N)$ is solvable. Thus, $\phi$ shuffles the normal subgroups in the intersection, and so sends $(U^p)^s$ to itself.
Since $(U^p)^s$ is characteristic in $U^p$ and $U^p$ is normal in $G$, then $(U^p)^s$ is normal in $G$.
Now notice that $G/U^s$ is solvable, and $U^s/(U^p)^s$ is solvable. Therefore, $G/(U^p)^s$ is also solvable, since it fits in an exact sequence
$$1 \longrightarrow \frac{U^s}{(U^p)^s} \longrightarrow \frac{G}{(U^p)^s} \longrightarrow \frac{G}{U^s}\longrightarrow 1,$$
and solvable groups are closed under extensions.
But that means that $U^s$, the smallest normal subgroup of $G$ such that the quotient is solvable, is contained in $(U^p)^s$, which is a normal subgroup of $G$ such that the quotient is solvable. That is, we have $U^s\leq (U^p)^s$. Since we already knew that $(U^p)^s\leq U^s$, we have equality.

Looking carefully at the argument above, we make use of the following facts:

*

*The class of $p$-groups is a pseudovariety (closed under subgroups, homomorphic images, and finite direct products); this ensures that $U^p$ is well-defined, and characteristic in $U$.


*The class of solvable groups is a pseudovariety; this ensures that $U^s$ is a well-defined characteristic subgroup of $U$, and that $U^p/U^s$ is solvable.


*Every $p$-group is solvable. This ensures that $U^s\leq U^p$.


*The class of solvable groups is closed under extensions. This ensures that $G/(U^p)^s$ is solvable, since it is an extension of $U^s/(U^p)^s$ by $G/U^s$. (Caveat: some people would call it an extension of $G/U^s$ by $U^s/(U^p)^s$ instead; see the extended discussion on this here.
Given two classes of groups, $\mathfrak{A}$ and $\mathfrak{B}$, closed under isomorphisms (that is, if $G\in\mathfrak{A}$ and $H\cong G$, then $H\in\mathfrak{A}$; similar with $\mathfrak{B}$); we define their product $\mathfrak{AB}$ to be the class of all groups $G$ that have a normal subgroup $N$ such that $N\in\mathfrak{A}$ and $G/N\in\mathfrak{B}$.
We have:
Theorem. Let $\mathfrak{P}$ and $\mathfrak{Q}$ be two pseudovarieties such that $\mathfrak{Q}\subseteq \mathfrak{P}$, and $\mathfrak{PP}\subseteq \mathfrak{P}$. Given a finite group $G$, let $\mathfrak{P}(G)$ (resp. $\mathfrak{Q}(G)$) be the least normal subgroup of $G$ such that $G/\mathfrak{P}(G)\in \mathfrak{P}$ (resp. $G/\mathfrak{Q}(G)\in\mathfrak{Q}$). Then $\mathfrak{P}(\mathfrak{Q}(G))=\mathfrak{P}(G)$.
Proof. Since $\mathfrak{Q}\subseteq \mathfrak{P}$, we have $\mathfrak{P}(G)\leq \mathfrak{Q}(G)$.
Since $\mathfrak{Q}(G)/\mathfrak{P}(G)$ is a subgroup of $G/\mathfrak{P}(G)$, and $\mathfrak{P}$ is closed under subgroups, we have $\mathfrak{P}(\mathfrak{Q}(G))\leq \mathfrak{P}(G)$.
And then we have that $\mathfrak{P}(G)/\mathfrak{P}(\mathfrak{Q}(G)) \leq \mathfrak{Q}(G)/\mathfrak{P}(\mathfrak{Q}(G))$. Therefore, $\mathfrak{P}(G)/\mathfrak{P}(\mathfrak{Q}(G)) \in\mathfrak{P}$.
Now we have a group extension given by
$$1 \longrightarrow \frac{\mathfrak{P}(G)}{\mathfrak{P}(\mathfrak{Q}(G))} \longrightarrow \frac{G}{\mathfrak{P}(\mathfrak{Q}(G))} \longrightarrow
\frac{G}{\mathfrak{P}(G)}\longrightarrow 1.$$
Since both the first and last terms of the sequence lie in $\mathfrak{P}$, the middle term lies in $\mathfrak{PP}\subseteq \mathfrak{P}$. Therefore, $\mathfrak{P}(G)\leq \mathfrak{P}(\mathfrak{Q}(G))$. As we already had the other inclusion, we conclude that $\mathfrak{P}(G)=\mathfrak{P}(\mathfrak{Q}(G))$, as claimed. $\Box$
To see that we do need the condition $\mathfrak{PP}\subseteq\mathfrak{P}$, we consider the case where $\mathfrak{P}$ is the class of nilpotent groups, and $\mathfrak{Q}$ is the class of $2$-groups. They are both pseudovarieties, $\mathfrak{Q}\subseteq\mathfrak{P}$, but in general an extension of a nilpotent group by a nilpotent group need not be nilpotent (for example, $S_3$ is an extension of $A_3$ by $C_2$).
Consider $G=S_3$. Then $\mathfrak{Q}(G) = A_3$, and since this is already abelian (hence nilpotent), we have $\mathfrak{P}(\mathfrak{Q}(G))=\{1\}$. On the other hand, $\mathfrak{P}(G) = A_3$. So here we have $\mathfrak{P}(\mathfrak{Q}(G))$ properly contained in $\mathfrak{P}(G)$.
