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**princess snowwhite****Member**- Registered: 2012-11-06
- Posts: 29

let G be a group of order p^2 where p is a prime no. let x belongs to G. prove that {y belongs to G: xy=yx}=G

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**princess snowwhite****Member**- Registered: 2012-11-06
- Posts: 29

does anyone know how to solve it?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,603

Hi

Is that the whole text for the problem?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**princess snowwhite****Member**- Registered: 2012-11-06
- Posts: 29

yes

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,468

Beats me too. But I've posted the link off to my son. He may be willing to help.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,603

What operation are you using in "xy=yx"?

*Last edited by anonimnystefy (2012-11-28 04:19:56)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**princess snowwhite****Member**- Registered: 2012-11-06
- Posts: 29

I am using multiplication.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,468

I'm wondering if the following is any help.

Assume y exists with the property and z is another element of G, so that z = xy (=yx)

xyx = (xy)x = zx

xyx = x(yx) = xz

so z has the property.

And this is true for all z in G

So can you prove that y exists?

LATER EDIT: Yes? because y could be the identity.

So H = {y : xy=yx} has a member. If z = xy, then z is also a member. Keep combining elements in this way from G until you have generated all of G.

What worries me is "what has p got to do with it?" So I assume the above is flawed but why?

If you can tell me that, maybe together we can fill in the gaps.

EVEN LATER EDIT

I think I've answered my own question.

H may be a subset of G rather than all of G.

So I'm working on:

(i) Show H is a group.

(ii) Lagrange => order of H is 1, p or p^2

If 1, then the problem is trivial.

If p^2, then the problem is solved.

So I'm aiming to show that order p is not possible, (maybe by showing order(H) > p )

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,603

Hi Bob

The proof using z doesn't work if you take the base element to be the identity element, because you generate only x itself!

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,468

By base element, I'm assuming you mean x.

Then it's easy.

H = {y∈G : iy = yi}

All y in G have this property so H = G straight away.

I've got to this.

Let x ∈ G.

xx = xx so x ∈ H

x(xx) = (xx)x so xx ∈ H

and so on.

So the cyclic group generated by x is a subset of H

note: cyclic groups are abelian.

If x ≠ i (the identity) then this subgroup has order p.

So now I'm trying to find a 'z' that is not in this cyclic group, but is in H.

One such would be sufficient as that makes the order of H > p

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**scientia****Member**- Registered: 2009-11-13
- Posts: 222

It is a well-known result that every

Suppose

. Then and so is cyclic, say . Let . Then and for some integers . Thus and for some .Hence *x* and *y* commute. Thus *G* is Abelian and so

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**scientia****Member**- Registered: 2009-11-13
- Posts: 222

Argh!

I've just realized that the subgroup princess snowwhite is defining is not the centre of

However, note that

is a subgroup of . Hence if (as I proved above) then necessarily as well so no harm done.Offline

**princess snowwhite****Member**- Registered: 2012-11-06
- Posts: 29

thanks to you all

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