Let $G = \{1,x,y\}$ be a group with identity element $1$. Prove that $x^2 = y, y^2 = x,$ and $xy = 1.$ 
Let $G = \{1,x,y\}$ be a group with identity element $1$. Prove that $x^2 = y$, $y^2 = x$, and $xy = 1$.

Every time I try this I end up with pages of nonsense. Any help would be appreciated.
 A: Let's fill in the Cayley table! Recall that Cayley tables are latin squares meaning that every element of the group appears exactly once in every row and every column (not counting the row and column headers). Let's start with a blank canvas:
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & & & \\ \hline \color{blue}x & & & \\ \hline \color{blue}y & & & \\ \hline\end{array}$$
First, we have that $1$ is the identity. Thus $1 \cdot 1 = 1$, $1 \cdot x = x \cdot 1 = x$, and $1 \cdot y = y \cdot 1 = y$, so
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & & \\ \hline \color{blue}y & y & \color{red}{?} & \\ \hline\end{array}$$
What can we fill in for the red question mark$\color{red}?$ Note that $y$ already appears in the given row, so it's not $y$. Similarly, $x$ already appears in the given column, so it's not $x$. Thus, it must be $1$.
In more concrete terms, $y \cdot x$ cannot equal $y$, as
$$y \cdot x = y \implies y^{-1} \cdot y \cdot x = y^{-1} \cdot y \implies x = 1,$$
and similarly,
$$y \cdot x = x \implies y = 1.$$
Updating our Cayley table,
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & & \color{red}{?} \\ \hline \color{blue}y & y & 1 & \\ \hline\end{array}$$
Similar logic works again here
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & & 1 \\ \hline \color{blue}y & y & 1 & \\ \hline\end{array}$$
We can now finish the whole Cayley table by filling in the blanks. The second row/column is missing a $y$, and the third row/column is missing an $x$. Thus,
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}y \\ \hline \color{blue}1 & 1 & x & y \\ \hline \color{blue}x & x & y & 1 \\ \hline \color{blue}y & y & 1 & x \\ \hline\end{array}$$
We can now read off the fact that $x^2 = y$ and $y^2 = x$.
A: Since $G$ has order $3$, you know that $x^3=y^3 = 1$. The inverse of $x$ cannot be $1$, or else we would have $x=1$. The inverse of $x$ cannot be $x$ itself, because $x^3=1$ would force $x=1$ again. So the inverse of $x$ is $y$, and we have $xy=1$. That being said, multiply both sides of $x^3=1$ by $y$ to get $x^2 = y$. Multiply by $y$ again to get $x = y^2$.
A: $x^2\in\{1,x,y\}$.  However, the order of $x$ has to divide the order of the group (which is $3$), and the order of $x$ can't be $1$ because $x\ne1$, so the order of $x$ must be $3$ and therefore $x^2\ne1$.  Furthermore, $x^2\ne x$, because otherwise $x=1$.  Therefore, $x^2$ must be $y$.
A similar argument proves that $y^2$ must be $x$.
Finally, $xy\in\{1,x,y\}$, but $xy$ can't be $x$ or else $y=1$, and $xy$ can't be $y$ or else $x=1$, 
so $xy$ must be $1$.
A: There is only one group with 3 elements.  I suppose that the point of this exercise is to show that this is the case.
What are the rules of groups?
The group is closed under our operation.
The operation is associative.
There is an identity element
Every element has an inverse element.
Fist show that the group described above is a group.
Suppose, we could describe another group with these elements?  What other operation could there be?
The identity element can only do one thing.
Suppose $x^2 \ne y.$   We could try $x^2 = 1$
But if that is the case, it would imply that $y^2 = 1$
What is $xy$?  
Working this way, you should be able to force a contradiction.
However you try to alter the group operation, it will lead to contradictions. 
