Recent content by potmobius

How are orbits and cosets related? Are all orbits cosets? Are all cosets orbits? Also, what exactly are G-sets and G-equivariant sets?

I did consider this situation, but didn't at first think it would be problematic. But clearly, h(t) doesn't have to equal n. and now f' would be useful. But I'm still not sure how to do so. Any hints on how to proceed?

I now see the importance of f. Since g(n)≠n, g(n) = p for some 1≤p≤n-1. Similarly, h(n) = q for some 1≤q≤n-1. Since we have freedom to choose f, we can define it such that f(p) = q. Now, if f' is the identity (I still don't see its importance), then h(n) = (f°g°f')(n). Similarly, we can define...

As it is restricted (can only permute the first n-1 elements), I think it is in Sn-1. But I think that since f is in Sn-1, f(n) = n. But g(n) ≠ n, so (f°g)(n) = n, which is not what we want, as h(n) ≠ n. But I think that one of f and f' should be a transposition (perhaps of order 2?) and the...

I still don't see why both f anf f' are needed. But doesn't defining f(k) = h°g^-1(k) for all 1<=k<=n-1 work? If f' is really necessary, it can be the identity, so h = f°g°f'

Ok. Either h(n) = n or not. If h(n) = n, then h is in Sn-1, in which case I don't know what g should be, as g(n) not equal to n, and regardless of what f and f' are, fgf'(n) won't equal n, which is a problem (but maybe you just disregard g completely in this case). Suppose h(n) not equal to n...

I don't know how to find such f,f', but my first guess was to define f inverse as f' and f as h restricted to the set (1,...,n-1), but I don't know if that actually works. But the equation you gave reminds me of conjugates. Are you trying to prove that h and g are conjugates? Or am I completely...

I don't know of such theorems. Are you perchance hinting to the set of generators or the set of orbits of Sn? But I'm thinking that if you take an element in Sn, call it sigma_n that's not in Sn-1, then it must shift the position of the nth element in the set (1,...,n), since if it were in...

I want to show that the only subgroups of Sn (the symmetric group of n elements) containing Sn-1 are Sn and Sn-1. So essentially, all that's needed to be checked is that there is no subgroup of order greater than (n-1)!, the order of Sn-1 and less than n!, the order of Sn. I was first thinking...

Do you know any bad math jokes related to this?

Thanks! Now I feel really stupid for not considering Rank-Nullity before asking this...

1. Say you have a linear transform from A to B, and where A has a higher dimension than B. How do you show that the kernel of the transform has more than one element (i.e. 0)? Also, if B has a higher dimension than A, then how to show that the transform isn't surjective? 2. The attempt at a...

I was looking up how pi was derived, and found this: http://mathforum.org/library/drmath/view/52589.html Question: How were the quantities n*sin(180/n) and n*tan(180/n) found/derived?

8% scored higher (more nerdy), 0% scored the same, and 92% scored lower (less nerdy). What does this mean? Your nerdiness is: Supreme Nerd. Apply for a professorship at MIT now!. Yay! :D

Yes, that's exactly what I meant, and this clears it up! Thanks :)