Here is the question I want to solve:
Prove that if $|G| = 2907,$ then $G$ is not simple.
Here is my idea for the solution:
We know that since, by Dummit and foote, a simple group $G$ is a group in which the only normal subgroups are the trivial ones, namely $1$ and $G,$ So what we want to show here is that the given group $G$ has a normal subgroup i.e., we want to show that there is a unique Sylow $p$-subgroup i.e., $n_p = 1$ for some prime $p.$
Also, we have $|G|= 2907 = 3^{2} \times 17 \times 19$. And we know that if $p^k$ is the highest power of a prime $p$ dividing the order of a finite group $G$, then a subgroup of $G$ of order $p^k$ is called a Sylow $p$-subgroup of $G$. So, our group $G$ has $3$ Sylow $p$-subgroups which are:
- $P$ a Sylow $3$-subgroup of order $3^2.$
- $Q$ a Sylow $17$-subgroup of order $17.$
- $R$ a Sylow $19$-subgroup of order $19.$
We will start first with the highest prime number:
The number of $19$-Sylow subgroups that exists.
we have $$n_{19} \mid 3^2 \times 17 = 153 \quad \quad (1)$$
But we also know that $n_{19}\equiv 1 \pmod p$ which means that $$n_{19} = 1 + 19k \quad \quad (2)$$ Therefore $$n_{19} = 1 (k=0)$$ and $$n_{19} = 153 (k=8).$$
The number of $17$-Sylow subgroups that exists.
We have $$n_{17} \mid 3^2 \times 19 = 171 \quad \quad (1)$$
But we also know $n_{17}\equiv 1 \pmod p$ which means that $$n_{17} = 1 + 17k \quad \quad (2)$$ Therefore $n_{17} = 1 (k=0)$ and $n_{17} = 171 (k=8).$
But then I do not know how do we know the number of elements in a group knowing that two Sylow subgroups (namely $17$– sylow and $19$-sylow subgroups)that are intersecting trivially?
Could anyone help me in this please?
Best Answer
If $n_{17}=171$ and $n_{19}=153$, then the number of elements of order $17$ is $171\times 16=2736$ and the number of elements of order $19$ is $153 \times 18=2754$ and hence the total number of elements in the Group is more than $2907$ which is not possible. Hence one of them must be one which gives a normal subgroup. Done.