You are not logged in.
#1 2012-11-28 03:57:55
- princess snowwhite
- Member
Offline
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
#3 2012-11-29 01:03:04
- anonimnystefy
- Real Member
Offline
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
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#6 2012-11-29 03:06:40
- anonimnystefy
- Real Member
Offline
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
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#8 2012-11-29 19:47:37
- bob bundy
- Moderator
Offline
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
- Real Member
Offline
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
“A secret's worth depends on the people from whom it must be kept.” ― Carlos Ruiz Zafón
#10 2012-11-30 03:35:19
- bob bundy
- Moderator
Offline
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
- Full Member
Offline
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
- Full Member
Offline
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.
