Timeline for Representations of finite abelian groups

8 events

when toggle format	what		by	license	comment
Jan 17 at 21:39	comment	added	Miranda		@YCor: Ah, good point, thanks to all!
Jan 17 at 21:26	review	Close votes
Jan 25 at 3:02
Jan 17 at 21:08	comment	added	YCor		@Miranda for nonzero $f$ in this subalgebra, $f^{-1}$ is a polynomial in $f$ so is in the subalgebra too.
Jan 17 at 19:22	comment	added	Miranda		Ah, I see: the point is that since $G$ is abelian, the image of $G$ in $End(V)$ lies in $End_G(V)$, which is a division ring. You then argue that the subalgebra generated by the image of $G$ in $End_G(V)$ is a commutative division ring (I guess you have to show that it is closed under inverses?), and thus a field. It then follows from the standard fact that Simon quotes. Is this a correct summary?
Jan 17 at 19:16	comment	added	YCor		(To apply the previous comment you need to observe that the subalgebra generated by the image of $G$ is a field.)
Jan 17 at 19:14	comment	added	Simon Wadsley		You need the result that every finite subgroup of the multiplicative group of a field is cyclic. The easiest way to prove this is by using that a polynomial of degree $d$ has at most $d$ roots and that a finite abelian group is cyclic if and only if it's order is its exponent.
S Jan 17 at 19:02	review	First questions
Jan 17 at 23:51
S Jan 17 at 19:02	history	asked	Miranda	CC BY-SA 4.0