Timeline for Transfer-free proof that the power map is a homomorphism
Current License: CC BY-SA 3.0
6 events
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| Jan 5, 2013 at 11:07 | comment | added | Ralph | In order to stress what Martin and Yves are saying, let $n$ be an integer such that $g^n$ is central for all $g$. Then, in general, the $n$th power map is no hom (take $G$ the dihedral group of order 8 and $n=2$). Hence, "formal manipulations" alone, like $(xy)^n=x(yx)^nx^{-1}=(yx)^n$ don't suffice to show that your map is a hom. | |
| Jan 5, 2013 at 3:09 | comment | added | YCor | @Martin and Todd: there is no good notion of index in monoids, so I don't really see which analogous statement would make sense. Actually the notion of index itself in the statement makes a "purely formal algebraic" approach quite unclear (at least to me). | |
| Jan 5, 2013 at 3:06 | comment | added | Tom Goodwillie | What statement is wrong for monoids? How would you define $(G:C)$? | |
| Jan 5, 2013 at 2:56 | comment | added | Todd Leason | @Martin: Yes, my question is about such a completely different approach. Thanks for clarifying. I don't understand why the failure in monoids rules out a proof based on formal manipulations: Can't such manipulations make use of the existence of inverses ? | |
| Jan 5, 2013 at 2:05 | comment | added | Martin Brandenburg | One can write down a direct proof which basically encodes the construction and the homomorphism property of the transfer map; which will be a quite long calculation with coset representatives. I wonder if there is a completely different approach (probably not). It cannot be a purely formal algebraic manipulation since the statement is wrong for monoids. | |
| Jan 5, 2013 at 1:41 | history | asked | Todd Leason | CC BY-SA 3.0 |