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+29 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 (106k points)
retagged by | 2.2k views


$\textbf{*}$ $\textbf{x}$ $\textbf{y}$ $\textbf{x*y}$ $\textbf{e}$
$\textbf{x}$ $e$ $x*y$ $y$ $x$
$\textbf{y}$ $x*y$ $e$ $x$ $y$
$\textbf{x*y}$  $y$ $x$ $e$ $x*y$
$\textbf{e}$ $x$ $y$ $x*y$ $e$


4 Answers

+44 votes
Best answer
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 by Veteran (13.3k points)
edited ago by

If every element in group G is equals to its own inverse then Group G is abelian Group.


Let (G,*) is a group.

∴  if  a,b ∈ G  then  a−1,b−1 ∈ G //Bcoz in Group inverse exist for every element.

also if a*b ∈ G the (a * b)−1 ∈ G //Bcoz in Group inverse exist for every element.

Given that a=a−1 and b=b−1

so we can write (a * b)=(a*b)−1=b−1 * a−1=(b * a)

i.e, (G,*) is commutative. Hence (G,*) is abelian group.

Nice approach to tackle such problems.Thanks :-)

Also it is given that x*x=e

this means order of group is surely even( By lagrange's theorem)

Since, 3 elements of the group are confirm and they are x,y and e

The group must atleast 4 or 6 or 8...elements.

Now exactly how much?

That is well explained above
converse need not be true right?
@Sushmita,please state the converse statement
@Ayush Upadhyaya

Converse is: If G is an abelian group then then every element is inverse of itself.

This need not be true always.
+21 votes
$x * x=e \quad\Rightarrow \text{x is its own inverse.}$

$y * y=e \quad\Rightarrow \text{y is its own inverse.}$

$(x * y)*(x*y)=e \quad\Rightarrow \text{(x*y) is its own inverse.}$

$(y*x)*(y*x)=e \quad\Rightarrow \text{(y*x) is its own inverse.}$

also $x*x*e=e*e$ can be rewritten as follows,

$x*y*y*x=e*y*y*e=[\therefore y*y=e ]$

$(x*y)*(y*x)=e$ shows that $(x*y) \text{ and }(y*x)$

are each other's inverse and we already know that $(x*y) \text{ and }(y*x)$ are inverse of its own.

As per $(G,*)$ to be group any elenent shouls hsve only one inverse element (unique).

This process $x*y=y*x$ (is one element).

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

And so the answer is 4.
answered by Veteran (19.6k points)
edited by
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 ...?
+17 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 (30.5k points)
Nice explanation.
+8 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 (691 points)
    how did you prove each element is its inverse?

    You have just used elements given in the equation.Group may or may not have more elements

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