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+8 votes
There are two elements $x,\:y$ in a group $(G,*)$ 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
where $e$ is the identity element. The maximum number of elements in such a group is ____.
asked in Set Theory & Algebra by Veteran (73.2k points)  
retagged by | 793 views

4 Answers

+13 votes

And so the answer is 4.

answered by Veteran (18.8k points)  
tell the elements...
can you see the image?

I didn't get it.How you got the 4 elements as (x,y,e,xy).?
I know i'm wrong but shouldn't it be (x,y,yx,xy)???

Above method is understandable but isn't there some easy way to solve this question ...?
+9 votes

the elements which are visible in this group at the first site are = {x, y, e, x*y, y*x}.

we can try generating other elements like
but these will simplify out to 

Property of a Group is that There can be only one identity element in a group, and each element in a group has exactly one inverse element.

we try to obtain inverse of x*y :
x*x = e
x*x*y = e*y
x*xy*y = y*y
x*xy*y = e
x*x*xy*y = x*e
e*xy*y = x
xy*y*x = x*x
xy * yx = e

This shows us that x*y has another inverse which is y*x. But as per the properties of a Group an element has a unique inverse. We already know that x*y is its own inverse, this is given to us. This makes us to conclude that x*y and y*x are same elements. Since it is given that G is a group and x*y and y*x both are its elements. So to keep them as elements of a Group they must be the same.

Hence there are only 4 elements in the Group (G,*), which are = {x, y, e, x*y} or we can write this set G as = {x, y, e, y*x}

answered by Veteran (26.8k points)  
+6 votes

x* x=e, x is its own inverse 
y *y= e, y is its own inverse

(x *y)*( x* y)= e, x *y is its own inverse

(y* x)* (y* x)= e, y* x is its own inverse
also x* x* e= e*e can be rewritten as follows

x* y* y *x= e *y* y* e= e, (Since y *y= e)
(x* y)* (y* x)= e shows that (x *y) and (y *x)
are each other’s inverse and we already know that
(x *y) and (y* x) are inverse of its own.

As per (G,*) to be group any element should have
only one inverse element (unique)

This implies x *y= y* x (is one element)

So the elements of such group are 4 which are {x, y,e,x *y }.

    answered by Junior (541 points)  
    +1 vote
    It is given that:

    $x$ is its own inverse.
    $y$ is its own inverse.
    $x*y$ is its own inverse.
    $y*x$ is its own inverse.

    Now i will show you that $x*y$ and $y*x$ are essentially same.
     $x*y = x*e*y = x*(x*y*x*y)*y = (x*x)*y*x*(y*y) = e*y*x*e = y*x$  (Group is associative so I do not care about brackets.)

    This turns out to be abelian group. and $x*y$ is no different from $y*x$

    Up to this point i have 4 elements - $x$, $y$, $e$, $x*y$.   (G is abelian therefore $x*y$ is same as $y*x$)
    Now see if you can have a new element. It is given that every element is product of some numbers of $x$ and $y$.
    Lets try with $x$.
    $x*\circ$, what u would like to put next to $x$ ?
    If you put $x$ then there is no use and you have to start over again because of $x*x=e$ now you have to start all over again.
    Put y next to x. : $x*y$ (this element we already have, we want different element so try multiplying further.)
    $x*y*\circ$, obviously u cant put $y$, next to $x*y$ because it will be x again: $x*y*y=x*e=x$
    (you have to put alternate.)
    Put $x$, next to $x*y$: $x*y*x$.
    This is equal to $x*x*y$ because of commutative property. $x*y*x=x*x*y=y$.
    I showed you that, once you get $x*y$ using $x$, you can not get next element by multiplying into $x*y$ further. Because of commutive property it will be again $x$ or $y$.

    Similarly, if we start with $y$, we have the same issue.

    This concludes that we can not generate further element and only four element can be there at max.
    There is a theorem for abelian group: If every element is its own inverse then Group G is abelian. I am not sure if proof of that theorem relates to this problem somewhere, You can check it out. :)
    answered ago by Loyal (3.5k points)  
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