Question

Let G be a group of order 11 2 :1 32 . H ow m any 11 -sylow subgroups and 13 sylow subgroups are there in G ?

Answer 1

Solution:

$O(G)=11^{2} 13^{2}$

The number of 11 Sylow subgroups are of the form t=1+11 k

since t divides O(G)

$\Rightarrow t\ divides\ 11^{2} 13^{2} \\\Rightarrow 1+11 k\ divides\ 13^{2} \\\Rightarrow 1+11 k=1 \\\Rightarrow k=0$

Number of 11- Sylow subgroup =1

Since all 11-Sylow subgroups are conjugate and there is only one 11-Sylow subgroup implies the 11-Sylow subgroup is normal.

With the similar argument we can show that there is one 13-Sylow subgroup which is normal.