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#1 2012-11-28 03:57:55

princess snowwhite
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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

#2 2012-11-29 00:27:19

princess snowwhite
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Re: group

does anyone know how to solve it?

#3 2012-11-29 01:03:04

anonimnystefy
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Re: group

Hi

Is that the whole text for the problem?


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#4 2012-11-29 01:16:32

princess snowwhite
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Re: group

yes

#5 2012-11-29 02:29:10

bob bundy
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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

#6 2012-11-29 03:06:40

anonimnystefy
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Re: group

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

Last edited by anonimnystefy (2012-11-29 03:19:56)


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#7 2012-11-29 15:20:15

princess snowwhite
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Re: group

I am using multiplication.

#8 2012-11-29 19:47:37

bob bundy
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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

#9 2012-11-30 00:43:45

anonimnystefy
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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!


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

#10 2012-11-30 03:35:19

bob bundy
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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

#11 2012-11-30 23:42:00

scientia
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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.)

#12 2012-12-01 00:35:23

scientia
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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

#13 2012-12-01 04:30:55

princess snowwhite
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Re: group

thanks to you all smile

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