TITLE: $\pi$ -Hall normal subgroup is characteristic
QUESTION [6 upvotes]: I've this exercise on my textbook: "Show that if G is a group (not necessary soluble), a normal $\pi$ -Hall subgroup is characteristic." 

I've tried to resolve it in the following way. 
Let $\alpha$ an automorhism of G; if H$\ne$H$^\alpha$ then since H is normal HH$^\alpha$ is a subgroup of G containing properly H. But, for order reasons HH$^\alpha$ is another $\pi$ -Hall subgroup of G, that is absurd. 
My solution is correct? Thanks in advice and sorry for my bad English!

REPLY [4 votes]: Well your sentence "for order reasons"is a bit vague. Look at the factor group $H^{\alpha}H/H$. This is a $\pi'$-group since $|G:H|$ is a $\pi'$-number. But this factor group is also isomorphic (2nd isomorphism theorem) to $H^{\alpha}/(H \cap H^{\alpha})$, and this is a $\pi$-group. Hence either factor must be trivial, whence $H=H^{\alpha}$.