Not understanding a group theory exercise Here's the exercise:

Let $G$ be a group and $x,y \in G\setminus \{1\}$. Let $R$ be any rectangle in the body of the multiplication table of $G$ having $1$ as one of its vertices, $x$ a vertex in the same row as $1$, and $y$ as a vertex in the same column as $1$. Prove that the fourth vertex of $R$ depends only on $x$ and $y$ and not on the position of $1$.

I don't understand what I'm asked to do. What do they mean by "depends on $x$ and $y$"? What kind of dependence are they (probably) talking about? Same thing for $1$; what do they mean by saying that the fourth vertex doesn't depend on the position of $1$?
 A: When they say "depends on $x$ and $y$", you're meant to show that the value of the fourth vertex is a function of $x$ and $y$. In general, there are a lot of $1$'s on the multiplication table, and the goal is to show that, if the rectangle has the aforementioned properties, then the value of the fourth vertex does not depend on which $1$ you pick, or which $x$ and $y$ you pick in the multiplication table. In other words, it shouldn't matter how you pick the vertices of the rectangle, just on the values of $x$ and $y$.
Indeed, suppose the vertices are $(a,b), (c,b), (a,d), (c,d)$, such that $ab=1, ad=x, cb=y$. Then $cd = c\cdot 1\cdot d = c\cdot (ba)\cdot d = (cb)(ad) = yx$, which only depends on $x$ and $y$, and not on how the vertices are chosen. 
A: The statement can be reformulated as follows

Let $G$ be a group, with $a,b\in G$;
  
  
*
  
*suppose $ab=1$
  
*consider $d$ so that $ad=x$
  
*consider $c$ so that $cb=y$
  
  
  Compute $cd$

A: Well, it's possible that when you do this, you find out that the last entry in the rectangle is always equal to $x^2$. That "depends only on $x$ and $y$." I think a better way to express this would be "Can expressed in terms of only $x, y, 1,$ and the group operations". 
There's also the problem that "1" appears in every row and column of the multiplication table, so saying "$x$ a vertex in the same row as $1$" 
doesn't constrain $x$ at all. What's almost certainly meant is this: 

Let $R$ be any rectangle in the body of the multiplication table of
  $G$--- which we represent as an array $A = (a_{ij})$ with entries in
  $G$--- having $1$ as one of its vertices, say $a_{pq}$, and $x$ as
  another entry in row $p$, $y$ another entry in columnn $q$, and $z$ as
  the fourth corner of the rectangle. Express $z$ in terms of $x, y, 1$
  and the group operations.

A: Well, there are various possibilities:
1 . . . . x     x . . . . 1
. . . . . .     . . . . . .
y . . . . z     z . . . . y

etc. I think the question is just asking you to show that $z$ has the same value in all cases.
