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#1 2012-11-27 04:57:55

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

group

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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#2 2012-11-28 01:27:19

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

Re: group

does anyone know how to solve it?

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#3 2012-11-28 02:03:04

anonimnystefy
Real Member
From: The Foundation
Registered: 2011-05-23
Posts: 15,609

Re: group

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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#4 2012-11-28 02:16:32

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

Re: group

yes

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#5 2012-11-28 03:29:10

bob bundy
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Registered: 2010-06-20
Posts: 6,471

Re: group

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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#6 2012-11-28 04:06:40

anonimnystefy
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From: The Foundation
Registered: 2011-05-23
Posts: 15,609

Re: group

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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#7 2012-11-28 16:20:15

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

Re: group

I am using multiplication.

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#8 2012-11-28 20:47:37

bob bundy
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Registered: 2010-06-20
Posts: 6,471

Re: group

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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#9 2012-11-29 01:43:45

anonimnystefy
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From: The Foundation
Registered: 2011-05-23
Posts: 15,609

Re: group

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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#10 2012-11-29 04:35:19

bob bundy
Moderator
Registered: 2010-06-20
Posts: 6,471

Re: group

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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#11 2012-11-30 00:42:00

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

Re: group


It is a well-known result that every p-group has a nontrivial centre. (This can be proved by counting conjugacy classes.) Hence if
, p prime, either
or
by Lagrange.

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

. (Thus the case
is acutally impossible.)

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#12 2012-11-30 01:35:23

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

Re: group

Argh!


I've just realized that the subgroup princess snowwhite is defining is not the centre of G but the centralizer of a fixed element x in G:
. (This is a subgroup of G).

However, note that

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

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#13 2012-11-30 05:30:55

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

Re: group

thanks to you all smile

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