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

In article <14506877.1221417997825.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
steenrod <spingroup@xxxxxxxxx> wrote:
In article
<5320186.1221376368842.JavaMail.jakarta@xxxxxxxxxxxxxx
orum.org>,
steenrod <spingroup@xxxxxxxxx> wrote:
Suppose a group G is a product of two of its
subgroups:
G = AB.

Let A' be a subgroup of A of index n, and
let B' be a subgroup of B of index m.
We want to show that the subgroup H = <A', B'> has
index at most mn.


Here's the proof:

Write G as a finite union
of left cosets of A' and right cosets of B';
since A'B' is contained in H, then G is just a union
of
right cosets of conjugates of H.

By the covering lemma, some conjugate of H has
finite index, hence so
does H. Moreover, the index of the core of H also
has finite index in
G, so it's enough to assume that G is finite.

If G is finite, then certainly

|G| = |A||B|/|A/\B| =< |A||B|/|A' /\ B'| =< |H|mn.

-----------


Question! Why is it enough to assume in this proof
that G is finite,
relying on the observation that G/H_G is finite?
This is the part
that's a little confusing to me.

You've been running into this particular problem
several times all
throughout this last week. This is simply the lattice
isomorphism
theorem, that yields a one-to-one, normality
preserving, inclusion
preserving correspondence between subgroups of G/K
and subgroups of G
that contain K, via the map that identifies the
subgroup H of G that
contains K with the subgroup H/K = {hK : h in H} of
G/K.


[...]

But the thing about the proof is that it alludes to the G-core
(again) of a finite-index subgroup H,

Yes: the fact that any subgroup of finite index contains a NORMAL
subgroup of finite index.

and jumps to the finite case without hesitation.

Because the problem is all about the index of H in G, and if you
already know that the index of H in G is finite, then the lattice
isomorphism theorem lets you go to the finite case without hesitation
by considering the index of H/K in G/K instead.

So, we know that G contains a finite-index subgroup H,
and we want to show that its index is at most mn (in G).

Great, so we know that, if K is the core of H in G,
then G/K is finite, since G acts on the set of left cosets of H in G, etc.

Is the proof trying to say that G/K contains H/K
as a subgroup of index at most mn?

Think about the cosets. The cosets of H/K in G/K are the sets of the
form

{ (gK)HK : g in G }

which is exactly the same as the sets

{ gK H : g in G}

(because K is contained in H)

which is exactly the same as the sets

{ gH : g in G}

because K is contained in K.

So there is a one-to-one correspondence between the cosets of H in G
and the cosets of H/K in G/K. This is all part of the lattice
isomoprhism theorem. That tells you that the index of H in G is
exactly the same as the index of H/K in G/K, so all you have to do is
work out the case when G is finite to get the case you are looking at.


--
======================================================================
"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

.

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