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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.

Shaun
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dedekind1
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  • The first problem: is it even clear that if $U$ and $V$ satisfy $P$ that $U\cap V$ must? Generally, no, and specifically, I think it's no for your $p$-group issue. – Randall Aug 02 '22 at 14:08
  • Think about property $P$ being "the quotient is cyclic." Apply your supposed universal definition to $U = \mathbb{Z}_2 \times \mathbb{Z}_2$. Then the subgroups $\mathbb{Z}_2 \times {0}$ and its vice versa contribute to $U(P)$, but their intersection does not. (In this case, there is no unique smallets $U(P)$ yielding a cyclic quotient.) – Randall Aug 02 '22 at 14:28
  • $|H\cap K|\cdot |HK|=|H|\cdot |K|$. Rearrange and multiply $|G|^2$ to both sides to obtain a formula for $|G:H\cap K|$. This shows that if $|G:H|$ and $|G:K|$ are not divisible by $r$, neither is $|G:H\cap K|$. The subgroup $U(\mathscr P)$ is called the $\mathscr P$-residual of $G$. Usually the quotient by it does not have property $\mathscr P$ for most properties $\mathscr P$. – David A. Craven Aug 02 '22 at 14:44
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    I have no idea what you mean by "all primes present in the Sylow decomposition of $U$", but the statement is definitely true for all primes $p$. If $p$ does not divide $|U|$, then $U^p=U$. – Derek Holt Aug 02 '22 at 14:54
  • This is proven with the exact same argument as the one for the smallest normal subgroup such that the quotient is solvable. Your previous question – Arturo Magidin Aug 02 '22 at 15:47
  • And it is well defined for all $p$. If $p$ does not divide the order of $U$, then $U^p=U$. – Arturo Magidin Aug 02 '22 at 15:50

3 Answers3

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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.

Nicky Hekster
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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.

David A. Craven
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  • @dedekind1 explicitly stated that $U$ is finite ... residually finiteness only make sense for infinite groups. Anyhow, +1 from me. – Nicky Hekster Aug 02 '22 at 14:57
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    @NickyHekster This is true, and is why I distinguished between the finite and infinite cases. Your answer was posted while I was writing mine, and yours should be accepted as it answers the question as stated. I only posted mine because I thought it would add a bit more context and I was 2/3rds of the way through anyway. – David A. Craven Aug 02 '22 at 15:00
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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.

Arturo Magidin
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