Re: - Finite-index subgroup in a product of subgroups

In article <14780260.1221440869703.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
steenrod <spingroup@xxxxxxxxx> wrote:
In the finite case,

|G| <= mn |H|

is equivalent to

|G|/|H| <= mn

which is equivalent to

[G:H] <= mn.

So use ->that<- inequality as equivalent to what you
want. In the
infinite case, with K a finite index normal subgroup
of G contained in
H, you will have

[G:H] = [(G/K):(H/K)] <= mn.



so

[G/K : H/K] =< mn


is automatic because we already know that happens in the finite case?

That is, whenever G is finite, and [G : H] =< mn;

then we can just replace G by G/K and H by H/K

to conclude

[G/K : H/K] =< mn


?

Is it not clear from the syntax of the statement? What exactly is
giving you trouble, and why? What is it about "for all finite groups"
that makes you doubt that it applies to a particular finite group,
namely the one you have before you?

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

.