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Let $G$ be a group and $H$ a subgroup.
$H$ is subnormal if it exists a finite normal chain from $H$ to $G$.
$H$ is quasinormal if $HS=SH$ for all subgroup $S$ of $G$.

If $G$ is a finite group, then every quasinormal subgroup is subnormal.
What about the converse :

Question: Is there a subnormal subgroup which is not quasinormal ?

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Groups in which subnormal=quasinormal have been classified. They are called T-groups or "groups in which normality is a transitive relation". – Jack Schmidt Mar 2 '14 at 15:22

For an explicit example, take $\;H:=\{(1)\,,\,(12)(34)\}\le S_4\;$ . Then this subgroup is subnormal since

$$H\lhd \{(1)\,,\,(12)(34)\,,\,(13)(24)\,,\,(14)(23)\}\lhd A_4\lhd S_4$$

but it is not quasinormal since if we take $\;K:=\langle(123)\rangle=\{(1)\,,\,(123)\,,\,(132)\}\;$ then



$$KH=\{(1)\,,\,(12)(34)\,,\,(123)\,,\,(134)\,,\,(132)\,,\,(234)\}\neq HK$$

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Thank you for this explicit example. – Sebastien Palcoux Mar 2 '14 at 13:15
up vote 2 down vote accepted

Yes, let $K$ be a non-trivial finite group.
Let $\tau \in Aut(K \times K)$ defined by $\tau(k_1,k_2)=(k_2,k_1)$.

Then $H=K \times \{ e\} \triangleleft K \times K \triangleleft (K \times K) \rtimes_{\tau} \mathbb{Z}_2 = G$, but $H.\mathbb{Z}_2 \neq \mathbb{Z}_2.H $

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