$ G $ is $ p $-supersolvable group . $ Q \in \operatorname{Syl}_{2}(G^{\prime}) $. Show $ Q \unlhd G $.

Question

Let $ G $ is a finite $ p $-supersolvable group for odd prime numbers . Suppose $ Q \in \operatorname{Syl}_{2}(G^{\prime}) $. Now i'll show $ Q \unlhd G $.

Since $ G $ is $ p $-supersolvable, then $ G^{\prime} $ is $ p $-nilpotent. Since $ p \nmid \vert Q \vert $, then $ Q \unlhd G $ ?

Answer 1

Since $ Q \in Syl_{2}(G^{\prime}) $, then $ Q \unlhd G^{\prime} $. So $ Q $ chr $ G^{\prime} $, since $ Q $ is normal sylow subgroup of finite group $ G^{\prime} $. Now we have $ Q $ chr $ G^{\prime} $ and $ G^{\prime} \unlhd G $. So $ Q \unlhd G $.