The order of the group 
The order of the smallest possible non trivial group containing elements $x$ and $y$ such that
  $x^7 = y^2 = e $ and $ yx = x^4 y$  is
  
  
*
  
*1
  
*2
  
*7
  
*14
  

I am stuck on this problem. Can anyone help me please?
 A: Up to isomorphism, there is only one group of order 1, one group of order 2, one group of order 7, and two groups of order 14. Figure out (or look up) what those groups are. Which ones are non-trivial? Look for suitable elements $x$ and $y$ in those groups. You may want to start from the smallest group and work your way up.
A: Hint:


*

*Lagrange's theorem: For any finite group $G$, the order of every subgroup $H$ of $G$ divides the order of $G$.

*$yx = x^4y$; $y(yx) = x$ do the manipulations using $yx = x^4y$ again and again and you will get $x = x^m$ for some $m \le 7 $ use that fact
A: First solution - $x=y=e$ satisfies the relations. The smallest non-trivial group has order 2, and the relations can be satisfied within that group.
Now suppose we want $x$ and $y$ distinct (not stated in the question).
If $x$ and $y$ are both non-trivial (i.e. $\neq e$) then the first relation shows that the group must contain non-trivial elements of orders 2 and 7, and then Lagrange means that the order of such a group must be divisible by 14. [note we have not used the second relation or shown it is compatible with this conclusion]
So to get a non-trivial group of order less than 14, one of $x$ or $y$ must be the identity. If we set $y=e$ we see that $x^7=x^3=1$ so that $x=e$, which is not what we want. If we set $x=e$ then $y^2=e$ and this can be done in a group of order 2.
A: For any group of order $2, say~\{e,a\}$ taking $x=e,y=a$ satisfies both the equation.
