Thread: Sylows Theorems

Hi got this question where I need to use sylows theorems (or one of them) to determine the numbers of sylow 11-subgroups and sylow 109-subgroups for a group G of order 11990 ($\displaystyle =2*5*11*109$).

Where do I start with this. Which of sylows theorems do i use and ultimately is it possible to show that G is not simple.

Originally Posted by moolimanj

Hi got this question where I need to use sylows theorems (or one of them) to determine the numbers of sylow 11-subgroups and sylow 109-subgroups for a group G of order 11990 ($\displaystyle =2*5*11*109$).

Where do I start with this. Which of sylows theorems do i use and ultimately is it possible to show that G is not simple.

Let $\displaystyle n$ be number of Sylow $\displaystyle 11$-subgroups. Then $\displaystyle n\equiv 1(\bmod 11)$ and $\displaystyle n|11990$. The only possibility is $\displaystyle n=1$. Thus there is only one Sylow subgroup $\displaystyle P$. But $\displaystyle aPa^{-1}$ is also a Sylow subgroup since it is unique it means $\displaystyle P = aPa^{-1}$ thus $\displaystyle P$ is normal.

So, do I just show that the group is normal and use the same process for the 109 group?

But isn't 2 x 5 x 109 = 1090 $\displaystyle \equiv$ 1 (mod 11)

and

2 x 5 x 11 = 110 $\displaystyle \equiv$ 1 (mod 109)

Yeah - thats interesting. Does that imply that there are more than one sylow subgroup for each?

PerfectHacker - can you clarify your working/thinking please?