let X(2n) be the group whose presentation is :(adsbygoogle = window.adsbygoogle || []).push({});

( 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

____________

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

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# Homework Help: A question in groups and presentation

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