If $G$ is a group of order $35=7\cdot5$ and $n_7$ is the number of the Sylow 7-subgroups then there are just $n_7\cdot 6$ elements of order 7?
Answer:
Yes because
If not suppose there exist $g$ with order 7 then $\langle g\rangle$ has order 7 and it is a subgroup of G then it must be Sylow 7-subrgoup
Am I right?
This holds in general for an element of order p and sylow-p subgroups?
Best Answer
To complete your argument you need to show that if $a,b$ both have order $7$ and $a^i=b^j\neq e$, for some integers $i,j$ then $\langle a\rangle= \langle b\rangle$.
That is you need to show that distinct Sylow $7$-subgroups do not intersect non-trivially (the intersection is just $\{e\}$). This should be an easy exercise in modular arithmetic.
To answer your last question note that sometimes distinct Sylow subgroups intersect non-trivially.
Consider the symmetry group of a hexagon. This has a Sylow $2$-subgroup for each pair of orthogonal reflections, consisting of the reflections, the identity and a $180^\circ$ rotation.
Here there are $3$ Sylow $2$-subgroups, each containing $3$ elements of order $2$, but only $7$ elements of order $2$ in total.