Re: algebra with order.

Arturo Magidin <magidin@xxxxxxxxxxxxxxxxx> wrote:
mina_world <mina_world@xxxxxxxxxxx> wrote:
"Arturo Magidin" <magidin@xxxxxxxxxxxxxxxxx> wrote:
mina_world <mina_world@xxxxxxxxxxx> wrote:

let G be an abelian group and
let H and K be finite cyclic subgroups with |H|=r and |K|=s.

if H=<a>, K=<b>, |ab| = lcm(r,s).

Assuming |ab| means the order of the element ab, this is false.
Take H=K, b=a^{-1}.

maybe, i look like confusing.
i know that if r and s are relatively prime, then |ab|=lcm(r,s)

Yes; in fact, the conclusion holds with more generality than the
case where r and s are relatively prime, but this much is correct.

and G contains a cyclic subgroup of order lcm(r,s)

Yes. But that subgroup need not be generated by ab. It can always be
generated by an element which is equal to a product of a power of a by
a power of b. If you don't know how, try the following hint: try to
pick a power of a, a^n, and a power of b, b^m, such that the orders of
a^n and of b^m are relatively prime, and the lcm of those orders
equals the lcm of r and s.

This is a FAQ. For further details see my prior post (esp. its link)
http://google.com/groups?threadm=y8ziswesnyl.fsf%40nestle.ai.mit.edu

--Bill Dubuque
.

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