# p-Sylow question

• May 30th 2010, 09:45 PM
aabsdr
p-Sylow question
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.
• May 30th 2010, 11:22 PM
tonio
Quote:

Originally Posted by aabsdr
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.

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

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

Tonio
• May 31st 2010, 02:23 PM
aabsdr
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!!!!!!!!!!!!!!!!!!!!