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

In article <galo6d$28n2$1@xxxxxxxxxxxxxxxxxx>,
Arturo Magidin <magidin@xxxxxxxxxxxxxxxxx> wrote:
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)]

[...]

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.

This, by the way, is the smallest counterexample in which the map is
not the trivial map. But just taking f to be the trivial map and
letting H be any proper subgroup would show you that the statement is
definitely incorrect.

--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

.