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I'm working with a Andreas Dress's article and he says

For a finite group $U$, we define $U^s$ to be the (well defined!) minimal normal subgroup of $U$ with $U/U^s$ solvable.

i can't see why $U^s$ is well defined. What motivates me is to prove that

$G$ solvable $\iff$ $U^s \sim V^s \text{, for all } U,V \leq G$

where $H\sim K \iff H$ and $K$ conjugated subgroups of G.

Thanks

dedekind1
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1 Answers1

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Well defined:

Let $G$ be a finite group. Suppose that $N$ and $M$ are two subgroups such that $G/N$ and $G/M$ are solvable. Since the product of finitely many solvable groups is itself solvable, we know that $(G/N)\times (G/M)$ is solvable. The map $G\to (G/N)\times (G/M)$ given by $g\mapsto (gN,gM)$ has kernel $N\cap M$, so $G/(N\cap M)$ is isomorphic to a subgroup of a solvable group, hence is itself solvable.

It follows that the intersection $N=\cap N_i$ of finitely many normal subgroups $N_i$ such that $G/N_i$ is solvable is itself a normal subgroup of $G$ such that $G/N$ is solvable. Moreover, $G/G$ is solvable, so every group has normal subgroups such that the quotient is solvable.

Thus, for every finite group $G$, there is a least normal subgroup $N$ of $G$ such that $G/N$ is solvable, namely, $N=\cap\{ M\triangleleft G\mid G/M\text{ is solvable}\}$.

The statement given:

As to what you want to prove: if $G$ is solvable, then every subgroup is solvable, so $U^s=\{e\}$ for all $U\leq G$; hence $U^s=V^s$ always holds. Conversely, since $\{e\}^s=\{e\}$, if $U^s\sim V^s$ for all $U,V\leq G$, then $U^s$ is conjugate to $\{e\}$, hence trivial, for all $U$. In particular, $G^s\sim\{e\}$, so $G^s=\{e\}$, which means that $G^s=\{e\}$. Thus, $G$ is solvable.

Shaun
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Arturo Magidin
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  • Thanks for answer. The argument that the normal subgroup $N=\cap N_i$ satisfy $G/N$ solvable, when $N_i$ are normal subgroups such that $G/N_i$ are solvable, does works when the quantity of sugbroups is finite. Can be the case that $N=\cap { M \triangleleft G | G/M \text{ solvable} }$ does not satisfy that $G/N$ is solvable? – dedekind1 Jul 30 '22 at 13:26
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    The argument fails when $G$ is not assumed to be finite. It fails for nonabelian free groups for example. – Derek Holt Jul 30 '22 at 13:36
  • @dedekind1 It can fail in general because an arbitrary product of solvable groups need not be solvable, if the solvability lengths of the factors are unbounded. Derek Holt's example is one such. – Arturo Magidin Jul 30 '22 at 14:00
  • then in the case that $G$ is a finite group, each $G/N_i$ has bounded solvability lenght, and the arbitrary groups product $\prod G/N_i$ is solvable, and then the argument holds, not only in the finite factors case. Is that right? – dedekind1 Jul 30 '22 at 14:11
  • @dedekind1 any particular solvable groups has a definite, finite, solvability length. "Bounded" is inapplicable to a single group. If you have only finitely many factors, then the length of the factors is bounded by the maximum length. That's why a finite product of solvable groups is always solvable. But in an infinite product, if there is no maximum length, then the product is not solvable. – Arturo Magidin Jul 30 '22 at 15:17
  • sorry, what i'm trying to say is that: let be ${M \triangleleft G | G/M \text{ solvable} }={M_i : i\in I}$, and $N=\cap_{i \in I} M_i$.\ If we define $f: G \rightarrow \prod_{i\in I} M_i$, $g\mapsto (g/M_i){i\in I}$, then $N=ker(f)\triangleleft G$, and $G/N \cong Im(f) \leq \prod{i\in I} M_i$ So, if the index set $I$ is finite, then $\prod_{i\in I} M_i$ is solvable and hence $G/N$ too, as you says. But if the set $I$ is not finite, then $\prod_{i\in I} M_i$ it's not necessarily solvable, and your argument doesn't works. Then why we can say that $G^s$ is well defined? – dedekind1 Jul 30 '22 at 15:58
  • @dedekind1 in the case where $G$ is infinite, we cannot. The least such subgroup may exist for some groups, but in general it will not. This is well defined for finite groups, which is the situation at hand. The definition specifies finite. – Arturo Magidin Jul 30 '22 at 16:18
  • Sorry, it was my mistake. In my notation $I$ is finite because $G$ is finite ($G$ finite iff $G$ has only finite many subgroups, I guess). Thanks for all – dedekind1 Jul 30 '22 at 17:02
  • @dedekind1 Yes, a group is finite if and only if it has only finitely many subgroups. – Arturo Magidin Jul 30 '22 at 18:31