Suppose $G$ is a group of order 4. Show either $G$ is cyclic or $x^2=e$. I've figured out that if I know $G$ is not cyclic, then it for any $a \in G, o(a) \neq 4$ (or the order of any element in group $G$ is not 4).
I know ahead of time that the elements in the group ($\forall x \in G$) must have order $o(x)=k$ where $0 < k \leq 4$ where $k \in \mathbb{Z}$, so $k = 1, 2, 3, 4$.
If $G$ not cyclic, then we know $k \neq 4$ so we have $k = 1, 2, 3$ left.
I know how to show that $k \neq 3$ but I am not sure if I am supposed to do it for all possible $k$ because obviously if our order was higher (let's say $n$) then it would get messy. It ends up being in this case that $k = 2$ is good and $k = 1$ is trivial. I have looked up online and it says using Lagrange's Thm we know it has to be $k = 1, 2$ since $3 \nmid 4$, but I cannot use that theorem as we have not have learned it in class. How can I show that $k = 1$ or $2$ in another way?
However working from using the result that if we have $x \in G$ and $H = \langle a \rangle$ and knowing that $\left\vert{H}\right\vert=1$ or $2$, then $x^2=1$ for either case. Then I am done right?
What is another, cleaner, better way of trying to answer the question in the title?
I apologize if my formatting is poor since I am new at LaTeX and extraordinarily bad at algebra it seems. Thank you in advance.
 A: First of all, by definition the order of an element $x\in G$ is the smallest positive number $n$ such that $x^n=e$. So we might as well rule out $O(x)=0$. ($O(x)=0$ doesn't even make sense anyway...what is $x^0$ supposed to mean?)
If $O(x)=1$, this by definition means that $x^1 = x = e$. In this case $x^2=e$ is trivial.
If $O(x)=2$, then $x^2=e$; this is already what we want.
$O(x)$ cannot be $4$, for then $x$ generates a cyclic group of $4$ elements, but we know $G$ is not cyclic.
The last case is $O(x)=3$. You say you know how to show this cannot be the case; now you're done!
A: This is essentially showing that if a group is of order 4 and not cyclic, it is the Klein 4 group isomorphic to $C_2 \times C_2$ - all groups of order 4 are isomorphic to one of these groups.
So to prove this without Lagrange's Theorem, we can suppose $o(x)=3$, else $o(x)=1$ if $o(x) \neq$, so x=e and this is trivial. So if $o(x)=3$, the group can be written as $G={e,x,x^2,y}$. But y has no inverse, so this is a contradiction. Therefore $o(x)=2$
A: Here is another way to look at this question:
$G$ has $4$ elements; call the identity $e$ and the others $a,b,c$. In writing the multiplication table for $G$ (which is feasible since $|G|$ is small:), we must have
\begin{array}{|c|c|c|c|c|}
\hline
\star& e & a & b & c \\ \hline
e &e &a &b & c\\ \hline
a &a  & &\\ \hline
b &b  & &\\ \hline
c &c & & &\\\hline
\end{array}
Now, each row and column of the multiplication table of a group must contain each element of the group exactly once (this is a good exercise to prove, using uniqueness of identity and so on). If $a^2=b^2=c^2=e$, then this is a valid multiplication table. If not, then there must be an element which is not its own inverse; WLOG assume $a\ne a^{-1}$. Then either $a^2=b$, which implies $ab=c$ and $ac = e$. In this case, $G=\{e,a,a^2,a^3\}$. Otherwise, $a^2=c$, which implies that $ba=e$ and $ca=b$. In this case again $G=\{e,a,a^2,a^3\}$. Quod erat demonstrandum.
(The purpose of this tedious approach is to motivate why we use theorems and do not attempt to enumerate algebraic objects by brute force ;)
