Question

Let G be the group of order 11^2: 13^2.how many 11 sylow subgroup and 13 sylow subgroup are in G?

Answer 1

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

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.