p-Sylow question

Apr 2010
22
0
Let K be a normal subgroup of a fintire group G. Let S be a p-Sylow subgroup of G (p a prime divisor of |G|).
Prove that KS/K is a p-Sylow subgroup of G/K.
Proof/
So we know that K is a subgroup of KS and KS is a subgroup of G, so by the correspondence theroem KS/K is a subgroup of G/K. Since |KS/K|=|S|/|K intersect S|, then |KS/K|= p (where p is a prime).
Want to show that KS/K is a miaximal p-subgroup of G/K.
So it suffices to show that [G/K:KS/K] is relatively prime to p. From here is where I'am having trouble I have been trying to use the facts about S being a p-sylow subgroup of G, to get that S/K is a p-sylow subgroup of G/K to try to get my conclusion,but I have been getting no where doing this. Any suggestions would be a life saver.
 
Oct 2009
4,261
1,836
Let K be a normal subgroup of a fintire group G. Let S be a p-Sylow subgroup of G (p a prime divisor of |G|).
Prove that KS/K is a p-Sylow subgroup of G/K.
Proof/
So we know that K is a subgroup of KS and KS is a subgroup of G, so by the correspondence theroem KS/K is a subgroup of G/K. Since |KS/K|=|S|/|K intersect S|, then |KS/K|= p (where p is a prime).
Want to show that KS/K is a miaximal p-subgroup of G/K.
So it suffices to show that [G/K:KS/K] is relatively prime to p. From here is where I'am having trouble I have been trying to use the facts about S being a p-sylow subgroup of G, to get that S/K is a p-sylow subgroup of G/K to try to get my conclusion,but I have been getting no where doing this. Any suggestions would be a life saver.

\(\displaystyle [G/K:KS/K]=\frac{\frac{|G|}{|K|}}{\frac{|KS}{|K|}}\) \(\displaystyle =\frac{|G|}{|KS|}=\frac{|G||K\cap S|}{|K||S|}\) . If we put \(\displaystyle |G|=p^rm\,,\,\,(p,m)=1\) , we get:

\(\displaystyle [G/K:KS/K]=\frac{|G||K\cap S|}{|K||S|}=\frac{p^rm\cdot |K\cap S|}{|K|p^r}=\frac{m|K\cap S|}{|K|}\) , and now we just have to note that any power of \(\displaystyle p\) in \(\displaystyle |K|\) cancels with the same power of \(\displaystyle p\) in \(\displaystyle |K\cap S|\) (why??)

Tonio
 
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Apr 2010
22
0
any power of p in |K| cancels with the same power of P in |K intersect S| because K intersect S is a maximal P subgroup of K.
Thanks you sooooooo much!!!!!!!!!!!!!!!!!!!!