Smallest normal subgroup such that the factor group is a $p$-group

Question

I am working with an Andreas Dress's article, and he says

for a group $U$ (finite), the subgroup $U^p$ is the (well defined !) smallest normal subgroup of $U$ with $U/U^p$ a $p$-group.

I think that is not well defined for all $p$ prime, but for all primes present in the Sylow descomposition of $U$.

However, I define $$U^p=\cap \{H \triangleleft U \mid U/H \text{ is a }p\text{-group} \}.$$

It iss easy verify that it's indeed a normal subgroup of $U$, but why $U/U^p$ is a $p$-group?

Let $\mathscr{P}$ be a property of groups.

Is $$U(\mathscr{P}):=\cap \{H \triangleleft U \mid H \text{ satisfies } \mathscr{P} \}$$ always 'the smallest normal subgroup of $U$ satisfying $\mathscr{P}$'?

Thanks for read.

Answer 1

For your first question: $U/U^p$ can be embedded in the direct product of all $U/H$, satisfying $H \unlhd U$ and $|U:H|$ is a $p$-power (map each $u \in U$ to $uH$ for each $H$, yielding an injective homomorphism with kernel $U^p$). Hence $U/U^p$ is isomorphic to a subgroup of a $p$-group and hence itself a $p$-group.

For your second question: if the property $\mathscr{P}$ is inherited by quotients, direct products and subgroups, then the answer is YES. Examples are being abelian, nilpotent, supersolvable or solvable. It is a NO for being cyclic.

Answer 2

Let $U(\mathscr P)$ be as given in the question. As mentioned in my comment, this is usually referred to as the $\mathscr P$-residual of $G$. If the property $\mathscr P$ is preserved by (subgroups, quotients and) finite direct products then $G/U(\mathscr P)$ has property $\mathscr P$, and usually not if this is not the case. If $G$ is infinite, then one needs to allow infinite products.

So the cyclic and abelian residuals are both the derived subgroup, and the soluble residual is the last term in the derived series. Your property, of being a $p$-group, is denoted by $O^p(G)$ in the literature. One of the most common properties used is finite, and residually finite groups (i.e., where $U(\mathscr P)=1$ for that group and property) are of great interest in infinite group theory. (They are, more or less, the collection of groups that can be understood through their finite quotients.) Residually nilpotent groups are also of interest in the literature.

Answer 3

A variety of groups is a nonempty class of groups that is closed under subgroups, homomorphic images/quotients, and arbitrary direct products. Examples of varieties of groups are:

1. Abelian groups.
2. Groups of exponent $n$.
3. Groups of solvability length at most $k$.
4. Groups of nilpotency class at most $k$.
5. All groups.
6. Just the trivial group.

A pseudovariety of groups is a class of groups that is closed under subgroups, homomorphic images/quotients, and finite direct products. Examples of pseudovarieties of groups are:

1. Finite groups.
2. Solvable groups.
3. Nilpotent groups.
4. Any variety of groups.

If $\mathfrak{V}$ is a variety of groups, then for every group $G$ there is a least normal (in fact, fully invariant) subgroup $N$ of $G$ such that $G/N\in\mathfrak{V}$. This subgroup is called the *verbal subgroup of $G$ associated to $\mathfrak{V}$, and is often denoted $\mathfrak{V}(G)$. This can be established a number of ways, but to parallel the development for pseudovarieties and what you've done, we can do the following: let $$\mathfrak{V}(G) = \bigcap\{N\triangleleft G\mid G/N\in \mathfrak{V}\}.$$ Note that since $\mathfrak{V}$ is nonempty, it must contain the trivial group (if $H\in\mathfrak{V}$, then $H/H\in\mathfrak{V}$). So the set we are intersecting is nonempty: it always contains $G$. Also we have that $\mathfrak{V}(G)$ is the kernel of the morphism $$G\hookrightarrow \prod_{G/N\in\mathfrak{V}} G/N$$ induced by the projections. This gives an embedding of $G/\mathfrak{V}(G)$ into the product. This product is in $\mathfrak{V}$, because it is a product of groups in $\mathfrak{V}$. Since $G/\mathfrak{V}(G)$ is isomorphic to a subgroup of a group in $\mathfrak{V}$, it is itself in $\mathfrak{V}$, as desired.

The fact that it is fully invariant is a bit harder to do in this set-up, but it should be clear that the group is characteristic, since any automorphism of $G$ just shuffles the $N$.

Now, the argument above doesn't work for pseudovarieties, because a pseudovariety is not closed under arbitrary products, only finite products. But when $G$ is a finite group, then there are only finitely many subgroups $N$, so the argument goes through. Thus, for every pseudovariety $\mathfrak{P}$ and every finite group $G$, there exists a unique normal (in fact, characteristic) subgroup $\mathfrak{P}(G)$ such that $G/\mathfrak{P}(G)\in\mathfrak{P}$. The argument is exactly the same as the one above.