The number of elements x s.t. $x^2 \neq 1$ is even Assume that G is a finite group.  I am asked to show that the number of elements x s.t. $x^2 \neq 1$ is even. 
What exactly does $x^2\neq1$ mean?  
Not only am I have trouble visualizing this but I would like a detailed explanation including concepts/theorems involved.  this is coming from a student who is taking groups theory for the first time.  
 A: It simply means $x \cdot x$ is not the identity element (here, $\cdot$ denotes the group operation). That is, $x$ has order greater than $2$. As for the problem you wish to solve, try pairing each such $x$ with its inverse (conveniently not $x$ wink wink). You should have nothing left over.
A: $\;x^2\neq 1\;$ are elements that do not have order two in the group. 
Now, to prove what you need pair each element in the group with its inverse if possible: $\;(x,x^{-1})\;$ . You can't do the above with, for example, the unit $\;1\;$ and neither with any element s.t. $\;x\neq x^{-1}\iff x^2\neq 1\;$ .
Now, the number of elements in pairs is obviously even, so...
A: $1$ is the identity element in the group. This notation is often used when the group operation is written as multiplication, because multiplying by $1$ is the identity operation in familiar contexts. $x^2$ means $x\cdot x$ where $\cdot$ is the group operation.
$x^2=1$ therefore means that the element $x$ is either the identity $(x=1)$ or that $x$ has order $2$.
$x^2\neq 1$ is equivalent to the statement that the order of $x$ is greater than $2$.
To prove a set is even, you can show that it can be split into pairs. It may not be immediately obvious how to do this, because you have very little to work with. So you need to look at that little you know about the elements of the group, which is essentially the definition of a group - from this we know that there is an identity element and every element has an inverse. You are also told that the group is finite.

Note that it is a little misleading to say that $x^2\neq 1$ is the same as the order of $x$ is greater than $2$, because this is not a statement which generalises. It is simply convenient for this problem.
If we had $x^3\neq 1$ it would mean that the order of $x$ is not $1$ (the identity) or $3$. For $x^4\neq 1$ it would mean that $x$ does not have order $1,2,4$.
In fact $x^n\neq 1$ in general would mean that the order $d$ of $x$ is not a factor of $n$.
