# Prove that for k, the number of elements with prime order p, k = -1 (mod p)

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

• Welcome to math stack exchange. On this site, you are expected to show what you have tried, where you got stuck and some other context. Hint to $c)$ : Sylow's theorem. If you have solved $b)$ , $a)$ is an immediate consequence of $c)$ and $b)$ Jun 8 '20 at 8:52
• Is it important that you show $a)$ first ? And what apart from Cauchy's theorem are you allowed to use ? Jun 8 '20 at 9:12
• I don't have to solve a) first but I cant use Sylow's thm as it has not been taught yet. I may use Causy, Lagrange, Euler and Fermat's little thrm Jun 8 '20 at 9:22
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 $$p$$.