By Sylow theorem, we see the number of the $7$-sylow subgroup is $n_7$. Then $n_7=1$ (mod $7$) and $n_7$ divides $15$; thus $n_7=15$, but why do we have $6\cdot 15=90$ elements of order $7$? And just by using $n_7,n_5,n_3$, how to answer the following question:
Is there a simple group of order $105$?
Thank you.
Best Answer
Two distinct $7$-Sylow subgroups intersect at the identity element (think about Langrange's Theorem). Hence, if there are $15$ distinct $7$-Sylow subgroups, it follows that, in addition to the identity element, there are at least $15\cdot 6 = 90$ elements of order $7$. But then we only have $15=105-90$ elements to account for. You can now argue that either $n_{3}=1$ or $n_{5}=1$.
Now, whenever $n_{p}=1$, this means there exists exactly one $p$-Sylow subgroup. Using Sylow's Second Theorem, this unique $p$-Sylow subgroup must be normal. As a result, the underlying group cannot be simple.
EDIT: As you-sir-33433 remarks, the normality of $p$-Sylow subgroup simply follows from the fact that $H$ and $gHg^{-1}$ have same cardinality for a subgroup $H\subset G$, and for any $g\in G$. One does not need Sylow's second theorem for this.