Find the number of elements of order $p$ in group $G$, where $p$ is prime and $G$ has $m$ subgroups of order $p$.
Suppose G is a group, and p is prime. Then the number of elements of G of order p is multiple of p-1. is an equivalent problem to this, but it was closed (also I have a more specific question).
The problem
Suppose that a finite group $G$ has exactly $m$ subgroups of order $p$, where $p$ is a prime. Show that the number of elements of order $p$ in $G$ is $m(p-1)$.
My solution
Let $H_i, 1 \leq i \leq m$, be every subgroup with order $p$ in $G$.
Theorem:
If $G$ is a group whose order is a prime $p$, then it is isomorphic to $C_p$(the cyclic group of order $p$).
Thus every $H_i$ is isomorphic to $C_p$.
Every cyclic group contains its generating element (of order $p$) exactly once, thus there must be at least $m$ elements in $G$ which has order $p$.
And that's as close as I can get to a solution.
$\endgroup$ 52 Answers
$\begingroup$Two subgroups of order $p$ are either equal or have trivial intersection.
Every nontrivial element in a subgroup of order $p$ has order $p$.
Each subgroup of order $p$ has $p-1$ elements of order $p$.
Every element of order $p$ belongs to a subgroup of order $p$.
Therefore, the number of elements of order $p$ is $m(p-1)$.
$\endgroup$ $\begingroup$My solution is almost correct, however there is a fault in my reasoning:
A cyclic group of order $n$ doesn't have just 1 generator, it has $\phi(n)$ generators where $\phi(n)$ is Euler's phi function. Or in other words: The amount of numbers which are coprime to the order of the cyclic group is equal to the amount of generators for that cyclic group.
Since $p$ is prime we have $p-1$ such numbers, thus we have $m(p-1)$ such elements in $G$.
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