Classifying groups of order 4 Let $G$ be a group of order $4$ then either $ G \cong C_4$ or $G \cong C_2 \times C_2$
the proof my lecture gave goes as follows:
Let $x \in G$ then by Lagrange $\text{ord}(x)$ divides $|G| =4 $ so w ehave either
1) $\exists x \in G$ with $x^4 = 1$
or 2) $\forall x \in G, x^2 = 1$
He then goes on to prove the theorem. I already don't understand this step and how it splits to 2 cases.
We have the $\text{ord}(x) $ divides $4$ so either $\text{ord}(x) =1, 2, 4$ if $\text{ord}(x) =1$ then $ x = 1$, what group does this give us?
if $\text{ord}(x)  = 2$ why does this result in $"\forall x \in G, x^2 = 1"$ why is it $\forall x$ whereas in case 1) when $\text{ord}(x) =4$ we have $\exists x$ why is one for all and one is there exists?
 A: The point is basically about logic, not algebra.
As you have stated, every $x\in G$ has order $1,2$ or $4$.  So there are two possibilities:


*

*there is an element of order $4$;

*there is no element of order $4$.


In the second case, every element must have order different from $4$, and therefore order $1$ or $2$, because these are the only remaining possibilities when $4$ is eliminated.
Now (in the second case), if $x$ has order $2$ then $x^2$ equals $1$ by definition.  If $x$ has order $1$ then $x^1=1$, and so $x^2$ still equals $1$.
A: We should really pick a nonidentity $x \in G$. If we pick an $x$ and it just so happens that $o(x) = 1$, then $x$ must be the identity element. If we pick the identity element, we don't learn anything: Every group has one of those; it's no help distinguishing between two groups!
The two cases are really: 


*

*There exists some element whose order is $4$, or

*There is no element whose order is $4$.
In either case, the order of an element has to be $1, 2,$ or $4$, based on Lagrange's theorem. But in the second case, if nothing has order $4$, then everything has to have order $1$ or $2$. But there's a unique element of order $1$ in a given group, the identity element. Therefore, every nonidentity element must have order $2$, in the second case.
