Number of elements of order $7$ in a group If $G$ has at least $7$ elements of order $7$, then prove that $G$ has at least $12$ elements of order $7$.


I know that an element having order $7$ means that for any $a$ in $G$, $a^7=e$ where $e$ is the identity element. The inverses also have same order, so I thought maybe of the $7$ elements, $6$ had their inverse as well. However, then this would just mean there had to be $8$ elements, not $12$ for the $7$th element to have it's inverse.
If there is an element with order 7, we have a subgroup of the group which is isomorphic to the cyclic group with 7 elements, namely the subgroup generated by the element of order 7.

All elements in this subgroup also have order seven (by Lagrange's theorem) excepting the identity element. So there exists another element with order seven out of the subgroup. This element also generates a subgroup of order 7 which meets the other subgroup only in the identity element (because intersections of subgroups are subgroups and by Lagrange a group of order seven has only the trivial subgroups), so we get 5 more elements of order seven, namely the other elements in the second subgroup.

So all in all we have at least 12 elements of order 7.
If $a$ has order $7$, then $a^2,a^3,a^4,a^5,a^6$ have order $7$.
Now the last one, $b$, is such that it has a different cycle. So $b^2,b^3,b^4,b^5,b^6$ have order $7$.