Math question (permutation groups)

[Ultraworld]

Question about permutation groups.
(1 2 3 4 5) and (1 3 4 5 2) are conjugate in Sym(5). However they are not conjugate in Alt(5). Why not?

conjugate means there exist a g in G such that (1 2 3 4 5) = g(1 3 4 5 2)g^-1

Maybe too easy but Im doing this course on my own with a prof in another city. It is nice when I could talk about it with other people. Even if the questions are easy.

got the answer (edit: almost). See below

 

[Ultraworld]

come on. not the most difficult one.

 

[Truronian]

I may be able to answer in a few weeks when I've actually revised permutations... I'm sure the more experienced resident mathematicians missed this one.

 

[newfangle]

Do it by brute force.

If the first one is a, the second one is b, show that gag^-1 =/= b for all g in alt(5). There are probably better ways, but alt(5) only has what, 30 elements?

It also probably follows from the normal subgroups in Alt(5). I'll let you ponder what those maybe be for a while.

 

[Atticus]

Maybe you should give some definitions, some of us are too lazy to check them.

 

[Swedishguy]

Say x is the answer. Now, imagine you know what x means. And, there you have it.

 

[Leifmk]

Sorry, I used to know this stuff, but that was ten years ago.

 

[Ultraworld]

Sorry, I used to know this stuff, but that was ten years ago.

Doesn't surprise me at all cause quite some norwegian mathematicians contributed to group theory

 

[Ultraworld]

ANSWER

There is a well known trick. For every g in G holds

g^-1(1 2 3 4 5)g = (g(1) g(2) g(3) g(4) g(5))

Now find all g's such that

g^-1(1 2 3 4 5)g = (g(1) g(2) g(3) g(4) g(5)) = (1 3 4 5 2), (3 4 5 2 1), (4 5 2 1 3), (5 2 1 3 4), (2 1 3 4 5).

The complete solution set of this equetion is

T = {(2 3 4 5), (1 2), (1 3 5)(2 4), (1 5 4 3), (1 4)(2 5 3)}

However all those g in T are not in Alt(5). So they are not conjugate

 

[nihilistic]

Question about permutation groups.
(1 2 3 4 5) and (1 3 4 5 2) are conjugate in Sym(5). However they are not conjugate in Alt(5). Why not?

conjugate means there exist a g in G such that (1 2 3 4 5) = g(1 3 4 5 2)g^-1

Maybe too easy but Im doing this course on my own with a prof in another city. It is nice when I could talk about it with other people. Even if the questions are easy.

Ooh when did this came up?

Anyway the conjugate should be (1 2). Just rearrange the two elements as: (1 2 3 4 5) and (2 1 3 4 5) and you should see why quite clearly.

As for the second part, you can assume there is some x in Alt(5) that fits the description, demonstrate that (1 2) * x must commute with (1 2 3 4 5) while x * (1 2) must commute with (1 3 4 5 2) and try to proceed onto a contradiction.

Do it by brute force.

If the first one is a, the second one is b, show that gag^-1 =/= b for all g in alt(5). There are probably better ways, but alt(5) only has what, 30 elements?

It also probably follows from the normal subgroups in Alt(5). I'll let you ponder what those maybe be for a while.

Alt(n) has n!/2 elements. Alt(5) is a simple group, so the only normal groups it has are itself and the trivial group. Got you there.

 

[Ultraworld]

thanks a lot nihilistic. Reading your post i discovered i forgot something. Now it is fixed and the answer is correct.

 

[pboily]

Slightly off topic, if you are in need of a top notch book, I suggest Abstract Algebra, by Dummit and Foote.

900 pages of algebra, but it looks like a first year Statistics test. Clever presentation, even cleverer exercices: it covers pretty much everything from Modular Arithmetic to Character Theory, and it does Galois theory on the way there.

 

[Ultraworld]

Thanks for the suggestion. At the moment I got Lang Algebra but browsing through some other books won't harm.


some magma code
http://magma.maths.usyd.edu.au/calc/

G := Sym(5);
elem1 := G ! (1, 2, 3, 4, 5);
elem2 := G ! (1, 3, 4, 5, 2);
g := G ! (1, 4)(2, 5, 3);

(g^(-1) * elem1 * g) eq (elem2);