Question

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.

Answer 1

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. :(

Answer 2

hmm... let's see :)

Answer 3

Automorphisms are isomorphisms?

Answer 4

Yes

Answer 5

Maybe cyclic?

Answer 6

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

Answer 7

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?

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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)

Answer 12

How do we prove G is cyclic?

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

What does that tell us? :)

Answer 17

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!

Answer 18

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

Answer 19

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?

Answer 20

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.

Answer 21

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.

Answer 22

LOL okay, I'm convinced :)

Answer 23

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

Answer 24

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!

Answer 25

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

Answer 26

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

Answer 27

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>?

Answer 28

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

Answer 29

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?

Answer 30

map generates a subgroup?