Math Is Fun Forum

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

princess snowwhite

Member

Registered: 2012-11-06

Posts: 29

Re: group

does anyone know how to solve it?

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

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

princess snowwhite

Member

Registered: 2012-11-06

Posts: 29

Re: group

yes

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: group

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

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

princess snowwhite

Member

Registered: 2012-11-06

Posts: 29

Re: group

I am using multiplication.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

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

---

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

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

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

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

---

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
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

scientia

Member

Registered: 2009-11-13

Posts: 224

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.)

scientia

Member

Registered: 2009-11-13

Posts: 224

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.

princess snowwhite

Member

Registered: 2012-11-06

Posts: 29

Re: group

thanks to you all