Question

There are two elements $x,y$ in a group $(G,\ast$) such that every element in the group can be written as a product of some number of $x$'s and $y$'s in some order. It is known that
$x\ast x=y\ast y=x\ast y\ast x\ast y=y\ast x\ast y\ast x=e$ where $e$ is the identity element. The maximum number of elements in such a group is

• A. 4

Answer 1

The correct option is A 4
(i) $e$ is identity element
(ii) $x\ast x=e$, so $x=x^{-1}$
(iii) $y\ast y=e$, so $y=y^{-1}$
(iv) $(x\ast y)\ast (x\ast y)=e$, so $x\ast y=(x\ast y{)}^{-1}\dots \dots (i)$
and $(y\ast x)\ast (y\ast x)=e,$ so $y\ast x=(y\ast x{)}^{-1}$
Now $(x\ast y)\ast (y\ast x)=x\ast (y\ast y)\ast x=x\ast e\ast x$
$=x\ast x=e$
So, $(x\ast y{)}^{-1}=(y\ast x)\dots \dots (ii)$
From (i) and (ii) we get $x\ast y\ast =y\ast x$
There are only 4 distinct elements possible in this group
1. $e$
2. $x$
3. $y$
4. $xy$
All other combinations are equal to one of these four as can be seen below:
$yx=xy$ (already proved)
$xxx=xe=x$
$xyy=xe=x$
$xxy=ey=y$
$xyx=xxy=y$
and so on....
So the group is $G=\{e,x,y,x\ast y\}$
$\Rightarrow |G|=4$