Prove x^2=1 if and only if the order of x is 1 or 2

[xsw001]

G is a group. Let x be an element of G.
Prove x^2=1 if and only if the order of x is 1 or 2.

How do I approach this problem?

I know since G is a group, all the elements in there have the following four properties:
1) Closure: a, b in G => a*b in G
2) Associative: (a*b)*c=a*(b*c)
3) Unique identity (e) exists: a*e=e*a=a
4) Unique inverse exists: a*a^(-1)=a^(-1)*a=e

 

[spamiam]

Recall the definition of the order of an element:

The order of an element x is the smallest positive integer n such that xⁿ = 1.

Can you find all the elements of order 1 in G?

 

[xsw001]

So assume x^2=1,
By definition of order of element if 2 is the smallest positive integer then order of x is 2.
Otherwise 2 is not the smallest, then 1 would be the only smallest positive integer, then the order of element x would be 1.
Does that look like a valid proof?

 

[spamiam]

Looks good to me!

And just to note, the only element of order 1 in a group is the identity.

 

[VietDao29]

So assume x^2=1,
By definition of order of element if 2 is the smallest positive integer then order of x is 2.
Otherwise 2 is not the smallest, then 1 would be the only smallest positive integer, then the order of element x would be 1.
Does that look like a valid proof?

This looks good. You can also reason like this, which is nearly the same:
Since we have: x² = 1.
And by the definition of 'order of an element', we have the oder of an element x is the smallest positive integer n, such that: xⁿ = 1. Which in turns means that it must be smaller or equal to 2 (since x² = 1), or in mathematical terms n \le 2 \Rightarrow n \in \{ 1; 2 \}.

However, you should note that this is a 'if and only if' statement. You're missing the proof for the \Leftarrow part! This should be pretty straight-forward. Let's see if you can tackle it yourself. :)