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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

miracle173
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Sofia
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    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 $b)$ – Peter Jun 08 '20 at 08:52
  • Is it important that you show $a)$ first ? And what apart from Cauchy's theorem are you allowed to use ? – Peter Jun 08 '20 at 09:12
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    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 – Sofia Jun 08 '20 at 09:22
  • @Sofia please add this to your question – miracle173 Jun 08 '20 at 09:42

1 Answers1

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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 $p$.

Nicky Hekster
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