Let p be a prime number and let G be a finite group whose order is divisible by p. Let k be the number of elements $x \in G$ of order p and let $l$ be the number of subgroups $ H \subseteq G $ of order p.
Prove:
a) $k \equiv -1$ (mod $p$)
b) $k = (p-1) \cdot l $
c) $ l \equiv 1$ (mod $p$)
So I have an idea about b)
If $H$ is a subgroup of order p then $H$ is cyclic.
Thus $\forall x \in H:\text{ord}(x) = 1\ \text{or}\ \text{ord}(x) = p$
So we have that $ k= (p-1) l$ as all elements in $\{ x\in G: \text{ord}(x) =p \}$ are all elements belong to one the subgroups $H$ excluding the identity
Is this correct?
Any hints for c) ?
(I think I can prove a) if I have proven c) and b))
I may use Cauchy, Lagrange, Euler and Fermat's little theorem
I urge you to read this entry, it is all explained there. From the proof (1959) of James McKay of Cauchy's Theorem (the existence of an element of prime power order $p$, if $p$ divides the order of $G$) (a) follows.
If there are $l$ subgroups of order $p$, then $l=1$ or any pair of different subgroups of order $p$ have trivial intersection (apply Lagrange's Theorem on the intersection, being a subgroup of each of the subgroups of order $p$).
Hence, leaving out the identity element, there are $l(p-1)$ elements of order $p$, yielding (b). Combining (a) and (b) gives $l \equiv 1$ mod $p$.