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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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

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

The knowledge of some things as a function of age is a delta function.

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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: 7,613

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

Bob

Children are not defined by school ...........The Fonz

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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

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

The knowledge of some things as a function of age is a delta function.

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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: 7,613

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

Children are not defined by school ...........The Fonz

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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

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

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,613

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

Children are not defined by school ...........The Fonz

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: 224

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: 224

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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