What is the Proof for Finite Group with Elements of Order p?

[MrsDebby]

Homework Statement

Suppose a finite group has exactly n elements of order p where p is prime. Prove that either n=0 or p divides n + 1.

Homework Equations

My professor says that this proof is similar to the proof of Lagrange's Theorem, in our Abstract Algebra book (Gallian).

The Attempt at a Solution

I am so lost with this question. It is a "special problem" we've been given to work on all semester. I have tried letting H represent a subgroup of the finite group, and using the elements of order p in the original group to form left cosets of H in the group. Not sure I really understand that. Not sure where that is getting me. I am not used to feeling so lost when tackling a problem.

Please...can anyone steer me in the right direction?

Thank you!

 

[Dick]

Start dealing with this by assuming there is an element x of order p. How many more elements of order p does that force the group to have? Assuming that those are all of the elements of order p, then what's the relation between n and p. Now generalize.

 

[MrsDebby]

thank you! I will work with that awhile...and let you know how it goes!

 

[MrsDebby]

I am sorry, I am still stuck. If x has order p, then x inverse must also have order p, and must also be in the group. then that is 2 elements...unless x is it's own inverse. So I think I am totally missing something here? HELP!
Thank you...

 

[Dick]

I am sorry, I am still stuck. If x has order p, then x inverse must also have order p, and must also be in the group. then that is 2 elements...unless x is it's own inverse. So I think I am totally missing something here? HELP!
Thank you...

x and x inverse aren't all. x generates a subgroup containing p elements. Try writing down a simple group of prime order like Z5. How many elements have order 5?

 

[Gigasoft]

What is \left(x^q\right)^p for some integer q? Is there a positive integer r smaller than p for which \left(x^q\right)^r=e might be true for some positive integer q less than p? If x is of order p, how many distinct elements that are powers of x exist? How many of those are of order p? For how many elements y that are powers of x, is x also a power of y, and vice versa? Then, if another element z that is not a power of x is of order p, how many distinct elements that are powers of z exist, and how many of those are of order p? It is seen that every subgroup of order p has only the element e in common, and the element e has order 1. This gives a general formula for the possible numbers of elements with order p.

 

[MrsDebby]

If x is of order p, there are p distinct elements of powers of x...right? <x> will have order p also. one of those will be e, so p-1 of them can have order p.
Z5 has 4 elements with order 5.

thinking...


sorry, I am usually not so dense..

 

[Gigasoft]

Yes, now consider the case of multiple subgroups of order p, such as Z_5\times Z_5:

e A A A A
B C D E F
B E C F D
B D F C E
B F E D C

 

[MrsDebby]

I'm sorry...I am still getting no where...

I do appreciate everyone's help, though. I will report if I make any progress. Going to talk to the professor tomorrow...this is so discouraging...