Does every group whose order is a power of .

[Jamin2112]

Does every group whose order is a power of ...

Homework Statement

Does every group whose order is a power of a prime p contain an element of order p?

Homework Equations

I'm trying to get away with only using Lagrange's Theorem and a corollary thereof.

The Attempt at a Solution

Seems like I'm almost there. How do I seal the deal? Thanks in advance, brahs.

 

[Dick]

You could use Cauchy's theorem, but that's overkill for this problem. Sure, finish your approach. If m is the order of some element g, then the group generated by g is a subgroup of order p^k for some k. And the subgroup generated by g is CYCLIC. Use that!

 

[Jamin2112]

You could use Cauchy's theorem, but that's overkill for this problem. Sure, finish your approach. If m is the order of some element g, then the group generated by g is a subgroup of order p^k for some k. And the subgroup generated by g is CYCLIC. Use that!

The subgroup generated by g is <g>={..., -g², -g, 1, g, g², ...}, which is the smallest subgroup of G that contains g, and has order p^k for some integer k≥0. Since p^k is not infinite, the elements of <g> are not distinct, so ... am I getting somewhere?

 

[Dick]

The subgroup generated by g is <g>={..., -g², -g, 1, g, g², ...}, which is the smallest subgroup of G that contains g, and has order p^k for some integer k≥0. Since p^k is not infinite, the elements of <g> are not distinct, so ... am I getting somewhere?

I think we are using multiplicative notation here. The distinct elements of the subgroup generated by g are {e,g,g^2,...g^(p^k-1)} since g^(p^k)=e. Now get somewhere with that. Isn't there an element of order p in there?

 

[Jamin2112]

I think we are using multiplicative notation here. The distinct elements of the subgroup generated by g are {e,g,g^2,...g^(p^k-1)} since g^(p^k)=e. Now get somewhere with that. Isn't there an element of order p in there?

*pssst* ... What's e?

 

[simmonj7]

e = indentity element of group G

 

[Jamin2112]

e = indentity element of group G

Got it.

Can you explain once more where we got {1, g, g², ..., g^pk-1}.

My book says "Let <x> be the cyclic group of a group G generated by an element x, and let S denote the set of integers k such that x^k=1.

[...]

Suppose that S is not the trivial subgroup. Then S = Zn for some positive integer n. The powers 1,x,x²,...,x^n-1 are the distinct elements of the subgroup <x>, and the order of <x> is n."

But I'm a little confused by it.

 

[Dick]

The subgroup generated by g is just the collection of all distinct powers of g. It's a subgroup since (g^a)(g^b)=g^(a+b). If the order of g is 5 then that's {1,g,g^2,g^3,g^4}. No use listing g^5 since that's 1 or g^6 since g^6=g^5*g=g. Now your case is order g equal to p^k.This is simple enough that I wouldn't spend a lot of time confusing yourself about it.

 

[Jamin2112]

This right?

 

[Dick]

This right?

Why do you think g^p has order p? And that's not generally correct. Proofs usually involve giving reasons. What's (g^p)^p? Why would that be 1?

 

[Jamin2112]

Why do you think g^p has order p? And that's not generally correct. Proofs usually involve giving reasons. What's (g^p)^p? Why would that be 1?

Never mind. So I'm looking for an element in {1, g, g², ..., g^pk-1} that has order p, i.e., the cyclic subgroup generated by that element this element has p elements. So if I let x be that element, the cyclic subgroup generated by x is {1, x, x², ..., x^p-1}. Am I right?

 

[Dick]

Never mind. So I'm looking for an element in {1, g, g², ..., g^pk-1} that has order p, i.e., the cyclic subgroup generated by that element this element has p elements. So if I let x be that element, the cyclic subgroup generated by x is {1, x, x², ..., x^p-1}. Am I right?

Yes. You want x^p=1. Which nonidentity element of {1, g, g², ..., g^pk-1} might have that property? Think about exponents ok? (g^a)^p=g^(ap) right? THINK!