A group is said to satisfy normalizer condition if its every proper subgroup $H$ is properly contained in its normalizer $N_G(H)$. Now my question is to show that in a finite group $G$ every subgroup is subnormal iff $G$ satisfies normalizer condition. One way I have proved that if every subgroup of $G$ is subnormal then $G$ satisfies normalizer condition but for the other way I am confused.
My attempt: Suppose $G$ satisfied normalizer condition then for any arbitrary subgroup $H$, we need to show that $H$ is term of some subnormal series of $G$. I have formed the series by this way
$H=H_0 \triangleleft N_G(H) \triangleleft N_G(N_G(H)) \cdots N_G(…(N_G(H))..)=G.$
Since $G$ is finite, therefore this series will terminate to $G$ after finite number of steps.
Is this correct proof?
Every subgroup of a group $G$ is subnormal $\iff$ $G$ satisfies normalizer condition.
abstract-algebrafinite-groupsgroup-theorysolution-verification
Best Answer
Every subgroup is normal its normalizer which is normal in its normalizer, etc. Since every proper subgroup is not equal to its normalizer that sequence of subgroups ends at the whole group. So every sugroup is subormal.