Re: algebra with order.

In article <e8gg41$pro$1@xxxxxxxxxxxxxxxx>,
mina_world <mina_world@xxxxxxxxxxx> wrote:
hello sir~

i know the next contents.

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}.

------------------------------------------------
if
G be an group.
|<a>| =r, |<b>| = s for some a, b in G.
ab = ba.

then, |ab| = lcm(r,s).

This is a complete nonsequitur.


--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
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Arturo Magidin
magidin@xxxxxxxxxxxxxxxxx

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