A question in groups and presentation

[Maths Lover]

let X(2n) be the group whose presentation is :
( x,y l x^n = y^2 = 1 , xy = yx^2 )

show that if n = 3k then X2n has order 6
and if (3,n)=1 then x satisfies the additional relation x=1 , in this case , deduce that X(2n) has order 2
note that :
x^3 = 1


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I tried to slove it and I think that I found the right solution and I will explain the main idea of it because the details is quite long , and I need to know if the main idea is right or not

the idea is :
first step ,
I suppose that S is a set : S= { 1 , x , y , x^2 , xy , yx }
then I proved that the elements of S are distinct " I mean that if a , b belongs to S then it's nessary that a=\= b "

then I proved that S = X(2n )
I did it as follows :
first I proved that every element of X(2n) is a product of finite elements if S
then I proved that any product of finite elements of S equals to some elements of S
the I deduced that every element of X(2n) belongs to S so X(2n) ⊂ S
but S ⊂ X(2n) because {x,y} generates X(2n) then {x,y} ⊂ X(2n) and every possible product of them also belongs to X(2n) because multipication is closed in groups

so X(2n) = S

but
lSl = the order of S = 6
so l X(2n) l = 6

at last , I show that if n is not equal to 3k then elements of S are not distinct so
l s l = 6 if n = 3k

is this idea true ??

the details is long

so I will omit it ,

and I wonder about your ideas to slove this problem ??
I find my one too long

thank you very much

 

[Asim Riaz]

Yes, your idea is correct. A simpler way of showing that the order of X(2n) is 6 when n = 3k is to note that since x^3 = 1, then for any element g in X(2n), we have g^6 = 1. This implies that the order of X(2n) is 6 or less. Since X(2n) is nontrivial, its order must be 6. If (3,n)=1, then x^n = 1. This implies that x = 1 and thus the group consists only of the identity element. Therefore, the order of X(2n) is 2 in this case.