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luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

group theory

G is a multiplicative group with identity element e. Given that G is commutative, prove that the set of all elements x for which x is it's own inverse forms a subgroup of G. (You should make clear where in your proof you use the fact that G is commutative.)

im kind of lost on getting it fully, anyone help?

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: group theory

Remember the four things to check: identity, closed under addition, closed under inverses, and associative.

identity and associative are both pretty much givens.

So let x and y be such that x^2 = y^2 = e.  What can you say about (xy)^2?

Now let x be such that x^2 = x.  What can you say about x-¹?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: group theory

don't you mean closed under multiplication its a multiplicative group, not an arithmetic group

I know the first hint, since its commutative, (xy)^n = x^ny^n, i had that one.

Actually you know what, it just occured to me where i couldn't take the leap. I was working through it, and seeing that for example, given 3 elements from G satisfying the conditions, the elements of the subgroup are e,x,y,z,xy,yz,xz,xyz

and i was having trouble being able to take the leap for when there are a non-finite number of elements taken from G. but ofcourse, i don't need to show it for every possibile combination, if i can show it for two of them, i.e. the x^2 = y^2 = e, then (xy)^2 = e, then i can say that whatever distinct element in the subgroup is generated from those taken from G, there all multiplications of eachover and so belong to G aswell, and however you take one element and multiply it with another, (or take one set of elements multiplied with a set of the other) it always gives either a new element not yet considered from G that is in G whos square is e, and inverse is itself or the identity. And so is valid.

Or rather, as i would answer it:

Any problems in that?

Last edited by luca-deltodesco (2008-06-03 11:17:06)

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The Beginning Of All Things To End.
The End Of All Things To Come.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: group theory

Any problems in that?

Only the typo:

and i was having trouble being able to take the leap for when there are a non-finite number of elements taken from G. but ofcourse, i don't need to show it for every possibile combination, if i can show it for two of them, i.e. the x^2 = y^2 = e, then (xy)^2 = e, then i can say that whatever distinct element in the subgroup is generated from those taken from G, there all multiplications of eachover and so belong to G aswell, and however you take one element and multiply it with another, (or take one set of elements multiplied with a set of the other) it always gives either a new element not yet considered from G that is in G whos square is e, and inverse is itself or the identity. And so is valid.

A hesitate to say that's right.  I think your overall idea is correct, but the way you worded it is a bit weird.  Instead of thinking  about it in terms of generating elements and such, you should instead think of it set theoretically, like you did in your latex.  All elements that belong to the subset in question, H, have a certain property.  No need to talk about what it is generated by, or even think about it.  Because if you can show it for two arbitrary elements of H, then it applies for any two elements of H.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."