Re: index of a subgroup and the index of any one of its conjugates (equality ?)

In article <5517685.1221452086658.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
steenrod <spingroup@xxxxxxxxx> wrote:
In article
<14079711.1221440175169.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org>,
steenrod <spingroup@xxxxxxxxx> wrote:

You lost the context again, by failing to quote.

I think it's enough to show that

[G : H] = [f(G) : f(H)]


for any homomorphism f : G -> G, yes?

Well, "enough" for what?

To settle the question I posed at the beginning?

And I'm expected to remember which of the many questions you posted
this thread is in, or what it is you are replying to?

Not everyone reads the newsgroup through the Math Forum or google. If
you do not provide enough context through quoting, I cannot know what
you are talking about.

And in any case, that much is false;


Really?

Yes. The claim that for ANY homomorphism f:G->G and any subgroup H of
G you have [G:H] = [f(G):f(H)] is most definitely false.

Here is a counterexample: let G = C_2 x C_2, the Klein 4 group. Let H
= {(0,0), (1,1)}. Let f:G->G be the map f(a,b) = (0,b).

Then [G:H] = 2. But f(G) = f(H) = {(0,0), (0,1)}, so [f(G):f(H)] = 1.

I tried defining a map


{left cosets of H in G} --> {left cosets of f(H) in f(G)}

via

xH --> f(x) f(H)


If xH = yH then x'y is in H (where x' := x^{-1}), so

f(x'y) is in f(H); equivalently,

f(x)' f(y) is in f(H), so that

f(x)f(H) = f(y)f(H), hence the map is well-defined.

A similar computation shows that it's injective

Cool. How do you show that it is injective with the specific example I
gave above? Note that (1,0)H maps to the same thing as (0,0)H.


it would
be true if you had either f one-to-one, or you had
ker(f) contained in
H, but otherwise, it can fail.


I would have never known.

What's wrong with the above argument, then?

Your "similar computation" to show the map is injective must be
wrong. But since you did not post it, I cannot tell you why.

I'm afraid I don't see anything wrong with it at this time.

I don't either, seeing as how you did not post it.

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