|G|=4. Prove the group is either cyclic or g^2=e

[mathmajor2013]

Let G be a group with |G|=4. Prove that either G is cyclic or for any x in G, x^2=e.

 

[disregardthat]

Consider and element x in G of maximum order. What possibilities are there, and what does each of them say about the group structure?

 

[mathmajor2013]

So the order of x is 2 or 4, right? Since the order must divide 4. If it's 2, the x^2=e, and so each element is its own inverse. If it's 4...I'm lost.

 

[disregardthat]

Correct. And if it's 4, what can you say about x, x^2,x^3 and x^4? Are any of them equal? What is x for this group?

 

[mathmajor2013]

Damn. They are all unique, and x^4=e. and x is the generator for the group yeah?

 

[mathmajor2013]

So the order of x is 2 or 4, right? Since the order must divide 4. If it's 2, the x^2=e, and so each element is its own inverse. If it's 4...I'm lost.

 

[Robert1986]

You are correct. The order of each element in the group must divide the order of the group. So, the order of each element must be 1,2 or 4. If there are no elements of order 4, then x^2=e for each x in the group (since the order of each element is 1 or 2). If there is an element of order 4, then this element is a generator, as you pointed out, and so the group is cyclic.

 

[lavinia]

Let G be a group with |G|=4. Prove that either G is cyclic or for any x in G, x^2=e.

a non identity element must have either order 4 or order 2