The mapping x to x^… - QuestionCove
OpenStudy (anonymous):

The mapping x to x^2 from a group G to itself is an automorphism. What can you say about G? Assume G is finite.

4 years ago
OpenStudy (anonymous):

I found out G is abelian, using the fact that the map is a homomorphism. And that no non-identity element has order 2. I cant proceed further. :(

4 years ago
OpenStudy (terenzreignz):

hmm... let's see :)

4 years ago
OpenStudy (terenzreignz):

Automorphisms are isomorphisms?

4 years ago
OpenStudy (anonymous):

Yes

4 years ago
OpenStudy (terenzreignz):

Maybe cyclic?

4 years ago
OpenStudy (terenzreignz):

Probably not, though, there's no element that may even remotely suggest a generator...

4 years ago
OpenStudy (anonymous):

I found out all cyclic groups of odd order have this automorphism through trial and error. How do I prove that G is such a group only?

4 years ago
OpenStudy (terenzreignz):

We don't even know if its order is finite yet... :D

4 years ago
OpenStudy (anonymous):

Oops. The question also assumes that G is finite. Its mentioned! My bad. Proceed using finite order group.

4 years ago
OpenStudy (terenzreignz):

Okay, so maybe it's cyclic... let's find out :D

4 years ago
OpenStudy (anonymous):

If its cyclic of even order, then it would have an non-identity element of order 2. So it cant be cyclic of even order. (since, injective so kernel is trivial)

4 years ago
OpenStudy (anonymous):

How do we prove G is cyclic?

4 years ago
OpenStudy (terenzreignz):

I don't even know if it is :) I'm just shooting random ideas about what we could possibly conclude about G :)

4 years ago
OpenStudy (terenzreignz):

Any other ideas about what we can conclude about the group? There are so few special groups :D

4 years ago
OpenStudy (anonymous):

What I am thinking is that, x is mapped to x^2 which is mapped to x^4 which is mapped to x^8 and so on. But this process must stop at some x^(2m) since G is finite.

4 years ago
OpenStudy (terenzreignz):

What does that tell us? :)

4 years ago
OpenStudy (anonymous):

That at some final stage, x^(2m)=x. That is, x^(2m-1)=1. So x has order (2m-1). So, every element in G has odd order. But I dont see how to correlate cyclicity and this!

4 years ago
OpenStudy (terenzreignz):

We don't know that x^2m = x do we? We just know that at some point, x^2m = x^2n where n < m

4 years ago
OpenStudy (anonymous):

Why? x is mapped to x^2 and x^2 to x^4 .... goes on... finally x^m mapped to x^2m, which cant be mapped further, cause there are no more powers of x as G is finite. So x^2m must be x. Am i correct?

4 years ago
OpenStudy (terenzreignz):

No... it just means we already passed x^2m at some point, but I don't see how it guarantees that it'd be x.

4 years ago
OpenStudy (anonymous):

Suppose all elements (powers of x) were never repeated until x^2m. Now x^2m cant be any other power of x than x. If it were, say, x^2m=x^2, then x^(2m-1) would have been the first element which was repeated, against our assumption.

4 years ago
OpenStudy (terenzreignz):

LOL okay, I'm convinced :)

4 years ago
OpenStudy (terenzreignz):

But still, that says nothing about whether or not it's cyclic... This stuff is hard without knowing what to aim for :D

4 years ago
OpenStudy (anonymous):

What we need to know now, is that are there any other elements other than powers of x in G? Is there a way to show that there isn't? Then we could prove that G is cyclic, hence killing the problem!

4 years ago
OpenStudy (terenzreignz):

Okay, let's start with that... suppose there is an element y such that $\Large x^n \ne y$ for any positive integer n

4 years ago
OpenStudy (terenzreignz):

Wait, no... it might be cyclic, but how do we know that x is the generator?

4 years ago
OpenStudy (anonymous):

Yeah, but the map generates a subgroup of G, that is (x , x^2 , x^4 , ... , 1). Its a odd order subgroup and all it's elements are even powers. Can I tell that its <x>?

4 years ago
OpenStudy (anonymous):

Suppose order of x is 5. Then (x, x^2, x^4, x^8=x^3, 1). See, its a cyclic group.

4 years ago
OpenStudy (anonymous):

Is the answer all abelian groups of odd order OR is it all cyclic groups of odd order? What is an example of abelian group of odd order which is not cyclic?

4 years ago
OpenStudy (terenzreignz):

map generates a subgroup?

4 years ago