2
$\begingroup$

Definition: A subgroup $H$ of a group $G$ is said to be normally embedded in $G$ if each Sylow $p$-subgroup of $H$ is a Sylow $p$-subgroup of some normal subgroup $N$ of $G$.

Definition: A group $G$ is a $T$-group if has a transitive relation among its subgroups i.e. all subnormal subgroups of $G$ are normal in $G$. The following is a classical result given by Gaschutz on the structure of the finite solvable $T$-groups

Theorem: Let $G$ be a finite solvable $T$-group. Then

(a) $G$ contains an abelian normal subgroup $L$.

(b) every subgroup of $G/L$ is normal in $G/L$


The following is adapted from an exercise in Finite Soluble Groups by Doerk and Klaus.

A finite group $G$ is a solvable $T$-group $\iff$ if every subgroup is normally embedded in $G$.

I have managed to show the sufficiency condition. For the necessary condition, fix a subgroup $H$ of $G$. I need to show that $H$ is normally embedded in $G$. Let $P$ be a Sylow $p$-subgroup of $H$ for some prime $p$. By the theorem, we know $G$ has a normal abelian subgroup $L$. Suppose that $p$ does not divide $|L|$. Then it is clear that $P$ is a Sylow $p$-subgroup of $PL$. Moreover, $PL/L$ is normal in $G/L$, and so $PL$ is normal in $G$. In this case, $H$ is normally embedded in $G$.

I'm not sure how to proceed for the case if $p$ divides $|L|$

$\endgroup$
0
$\begingroup$

Let $L = R \times Q$, where $R$ is a $p'$-group and $Q$ is a $p$-group. Now $PQ$ is a $p$-group, so $P$ is subnormal in $PQ$, and hence $PR$ is subnormal in $PL$, But $PL \unlhd G$, so $PR$ is subnormal and hence normal in $G$, with $P \in {\rm Syl}_p(PR)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.