Moderators: gmalivuk, Prelates, Moderators General
++$_ wrote:You can use one of the other criteria.
Hint: What are all the conjugates of (1 2)(3 4)?
++$_ wrote:Well, you don't need any advanced stuff here.
A subgroup K of G is normal iff all the conjugates of elements in K are also in K. (This is one of your criteria.)
So all you have to do is find out what the conjugates of the elements of K are.
The shortcut is that the set of conjugates in S_{n} of any permutation is easy to write down, so it is easy to check whether that is contained in K.
++$_ wrote:The shortcut is that the set of conjugates in S_{n} of any permutation is easy to write down, so it is easy to check whether that is contained in K.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Er, I don't agree with that last line. What is the inverse of (1 2 3)?gorcee wrote:So... the conjugates of (1 2)(3 4) would be what?
(1 2) (1 2)(3 4) (1 2)
(1 3) (1 2)(3 4) (1 3)
...
(1 2 3) (1 2)(3 4) (1 2 3)?
++$_ wrote:Er, I don't agree with that last line. What is the inverse of (1 2 3)?gorcee wrote:So... the conjugates of (1 2)(3 4) would be what?
(1 2) (1 2)(3 4) (1 2)
(1 3) (1 2)(3 4) (1 3)
...
(1 2 3) (1 2)(3 4) (1 2 3)?
++$_ wrote:Well, once you have discovered the general principle that all the conjugates of an element in S_{n} have the same ____________ as the original element, then you don't have to actually do any computation.
++$_ wrote:Well, it shouldn't be so hard to prove. If you can prove it for a single cycle, you can prove it for any permutation because if k_{1} and k_{2} are two disjoint cycles, then gk_{1}k_{2}g^{-1} = (gk_{1}g^{-1})(gk_{2}g^{-1}) (and these are disjoint cycles).
So you just have to check it for a single cycle.
++$_ wrote:Sorry, maybe I wasn't clear.
Basically, the key insight is that if g is any permutation and k = (a b c ... z), then gkg^{-1} is not just any cycle of the appropriate length, but specifically (g(a) g(b) g(c) ... g(z)). This is easy to prove -- just do it element by element.
Then suppose that (a b c ... z) and (A B C ... Z) are disjoint cycles. Using the previous result, the conjugate by g of (a b c ... z)(A B C ... Z) is (g(a) g(b) ... g(z))(g(A) g(B) ... g(Z)). These are still disjoint cycles, so the cycle structure of the permutation was again unchanged.
Hence, by induction on the number of cycles of the permutation, the cycle structure of any permutation is preserved by conjugation.
Users browsing this forum: Bing [Bot] and 1 guest