P-groups and orders of elements

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Homework Help Overview

The discussion revolves around proving that for a group H of order p^n, where p is a prime and n > 0, any element x not equal to the identity has some power that has order p. The subject area is group theory, specifically focusing on p-groups and the orders of their elements.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of Lagrange's theorem regarding the orders of elements in groups. There is a focus on finding specific powers of elements that yield an order of p, particularly when starting with elements of higher orders such as p^2 or p^r.

Discussion Status

The discussion is active, with participants questioning how to derive the order p from higher orders and examining specific cases. Some guidance has been offered regarding the properties of exponents and the relationship between different powers of elements, but no consensus has been reached on the general case.

Contextual Notes

Participants are working under the constraints of group theory and the properties of p-groups, with some uncertainty about the specific powers that yield the desired order. There is an ongoing exploration of the implications of the orders of elements and their powers.

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Homework Statement


Let p be a prime and let H be a group of order p^n, some n > 0. Prove that for any x not equal to 1 in H, some power of x has order p.

Homework Equations


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The Attempt at a Solution


I know that by lagrange, for any x in G, if x is not the identity, then x has an order p^r for some r>0. Also I know that there is some element of order p. But I don't know how to show that some power of every p^r = p, even though it seems almost intuitive.

Thanks very much!
 
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Suppose that x is not the identity and has order [itex]p^2[/itex]. Can you find a power of x that has order p? Now try to solve the general case where x has order [itex]p^r[/itex].

Petek
 
Petek said:
Suppose that x is not the identity and has order [itex]p^2[/itex]. Can you find a power of x that has order p?

Sorry, I don't even know which power of x would yield order p in the case of [itex]p^2[/itex]! Is it [itex]x^2[/itex]?
 
If you know x has order p^r for some r, then x^(p^r)=1. What's (x^(p^(r-1)))^p?
 
Dick said:
If you know x has order p^r for some r, then x^(p^r)=1. What's (x^(p^(r-1)))^p?

Using the example of the cyclic group of order 9, I found that (x^(p^(r-1)))^p is 1 as well. Is that correct?
 
kimberu said:
Using the example of the cyclic group of order 9, I found that (x^(p^(r-1)))^p is 1 as well. Is that correct?

Probably. But I meant it more generally. In any group, how does (x^(p^(r-1)))^p compare with x^(p^r)?
 
Dick said:
Probably. But I meant it more generally. In any group, how does (x^(p^(r-1)))^p compare with x^(p^r)?

(x^(p^(r-1)))^p is equal to x^(p^r)?

So, is (x^(p^(r-1))) always of order p?
 
Last edited:
kimberu said:
(x^(p^(r-1)))^p is equal to x^(p^r)?

So, is (x^(p^(r-1))) always of order p?

Yes, just use the rules of exponents. (x^n)^m=x^(nm).
 
Okay, so I have:

For any element x, |x|=p^r, x^p^r = 1 = (((x^p)^(r-1))^p) by the properties of exponents. Then x^p^(r-1) equals an element of order p.

Thank you so much!
 

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