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 Posted: Mon May 28, 2007 7:48 am Post subject: homomorphisms from a finite group to C* |
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if a finite group G has order N, is the number of homomorphisms from G
to multiplicative set of complex numbers (C*) also N? Why? |
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| Posted: Mon May 28, 2007 8:20 am Post subject: Re: homomorphisms from a finite group to C* |
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On May 27, 7:48 pm, oferl...@yahoo.com wrote:
| Quote: |
if a finite group G has order N, is the number of homomorphisms from G
to multiplicative set of complex numbers (C*) also N? Why?
|
Why not think about it for ten minutes before posting? Try some
examples. |
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Keith Ramsay
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| Posted: Mon May 28, 2007 8:57 am Post subject: Re: homomorphisms from a finite group to C* |
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On May 27, 10:42 pm, quasi <q...@null.set> wrote:
|Finally, prove it assuming G is nonabelian. Let K be the commutator
|subgroup of G. Show that every homomorphism from G/K to C corresponds
|to k distinct homomorphisms from G to C where k=|K|.
No, it only is true if G is Abelian. If G is not Abelian,
there are only |G/K|.
Keith Ramsay |
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quasi
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| Posted: Mon May 28, 2007 9:42 am Post subject: Re: homomorphisms from a finite group to C* |
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On 27 May 2007 19:48:53 -0700, oferlock@yahoo.com wrote:
| Quote: |
if a finite group G has order N, is the number of homomorphisms from G
to multiplicative set of complex numbers (C*) also N? Why?
|
Yes (I think).
Hints:
First prove it assuming G is cyclic. The elements of G would have to
map to what specific set of elements of C?
Next, prove it assuming G is abelian. Express G as a direct product of
cyclic groups.
Finally, prove it assuming G is nonabelian. Let K be the commutator
subgroup of G. Show that every homomorphism from G/K to C corresponds
to k distinct homomorphisms from G to C where k=|K|.
quasi |
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quasi
Guest
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| Posted: Mon May 28, 2007 10:01 am Post subject: Re: homomorphisms from a finite group to C* |
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On 27 May 2007 20:57:11 -0700, Keith Ramsay <kramsay@aol.com> wrote:
| Quote: |
On May 27, 10:42 pm, quasi <q...@null.set> wrote:
|Finally, prove it assuming G is nonabelian. Let K be the commutator
|subgroup of G. Show that every homomorphism from G/K to C corresponds
|to k distinct homomorphisms from G to C where k=|K|.
No, it only is true if G is Abelian. If G is not Abelian,
there are only |G/K|.
Keith Ramsay
|
Of course, I should have seen that.
For example, if K=G, then the only homomorphism from G to C is the
trivial one.
quasi |
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quasi
Guest
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| Posted: Mon May 28, 2007 10:15 am Post subject: Re: homomorphisms from a finite group to C* |
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On Mon, 28 May 2007 00:01:33 -0500, quasi <quasi@null.set> wrote:
| Quote: |
On 27 May 2007 20:57:11 -0700, Keith Ramsay <kramsay@aol.com> wrote:
On May 27, 10:42 pm, quasi <q...@null.set> wrote:
|Finally, prove it assuming G is nonabelian. Let K be the commutator
|subgroup of G. Show that every homomorphism from G/K to C corresponds
|to k distinct homomorphisms from G to C where k=|K|.
No, it only is true if G is Abelian. If G is not Abelian,
there are only |G/K|.
Keith Ramsay
Of course, I should have seen that.
For example, if K=G, then the only homomorphism from G to C is the
trivial one.
|
But wait, I take it back.
I'm not so sure you're right.
Every homomorphism from G to C factors through G/K, but there may be
more than one homomorphism from G onto G/K.
In fact, start with any homomorphism of G onto G/K and follow it with
an arbitrary automorphism of G/K to get a new homomorphism from G onto
G/K.
So I think the correct answer for the nonabelian case is m*|G/K| where
m=|aut(G/K)|.
quasi |
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Subluxian
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| Posted: Mon May 28, 2007 10:24 am Post subject: Re: homomorphisms from a finite group to C* |
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On May 27, 10:15 pm, quasi <q...@null.set> wrote:
| Quote: |
On Mon, 28 May 2007 00:01:33 -0500, quasi <q...@null.set> wrote:
On 27 May 2007 20:57:11 -0700, Keith Ramsay <kram...@aol.com> wrote:
On May 27, 10:42 pm, quasi <q...@null.set> wrote:
|Finally, prove it assuming G is nonabelian. Let K be the commutator
|subgroup of G. Show that every homomorphism from G/K to C corresponds
|to k distinct homomorphisms from G to C where k=|K|.
No, it only is true if G is Abelian. If G is not Abelian,
there are only |G/K|.
Keith Ramsay
Of course, I should have seen that.
For example, if K=G, then the only homomorphism from G to C is the
trivial one.
But wait, I take it back.
I'm not so sure you're right.
Every homomorphism from G to C factors through G/K, but there may be
more than one homomorphism from G onto G/K.
|
Yep.
| Quote: |
In fact, start with any homomorphism of G onto G/K and follow it with
an arbitrary automorphism of G/K to get a new homomorphism from G onto
G/K.
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Yes: If T in Hom(G,G/K) and A in Aut(G/K), then AT in Hom(G,G/K).
But it is not necessarily the case that AT is not equal to A'T for all
A, A' in Aut(G/K), T in Hom(G, G/K). So your claim that we always "get
a /new/ homomorphism" (emphasis mine) is false.
There is an obvious surjection Hom(G, G/K) x Aut(G/K) x Hom(G/K, C) ->
Hom(G,C); but we are interested in counting the elements of the range
of this surjection, not its domain.
Cheers - Chas |
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quasi
Guest
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| Posted: Mon May 28, 2007 11:01 am Post subject: Re: homomorphisms from a finite group to C* |
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On 27 May 2007 22:24:28 -0700, Subluxian <cbrown@cbrownsystems.com>
wrote:
| Quote: |
On May 27, 10:15 pm, quasi <q...@null.set> wrote:
On Mon, 28 May 2007 00:01:33 -0500, quasi <q...@null.set> wrote:
On 27 May 2007 20:57:11 -0700, Keith Ramsay <kram...@aol.com> wrote:
On May 27, 10:42 pm, quasi <q...@null.set> wrote:
|Finally, prove it assuming G is nonabelian. Let K be the commutator
|subgroup of G. Show that every homomorphism from G/K to C corresponds
|to k distinct homomorphisms from G to C where k=|K|.
No, it only is true if G is Abelian. If G is not Abelian,
there are only |G/K|.
Keith Ramsay
Of course, I should have seen that.
For example, if K=G, then the only homomorphism from G to C is the
trivial one.
But wait, I take it back.
I'm not so sure you're right.
Every homomorphism from G to C factors through G/K, but there may be
more than one homomorphism from G onto G/K.
Yep.
In fact, start with any homomorphism of G onto G/K and follow it with
an arbitrary automorphism of G/K to get a new homomorphism from G onto
G/K.
Yes: If T in Hom(G,G/K) and A in Aut(G/K), then AT in Hom(G,G/K).
But it is not necessarily the case that AT is not equal to A'T for all
A, A' in Aut(G/K), T in Hom(G, G/K). So your claim that we always "get
a /new/ homomorphism" (emphasis mine) is false.
|
But I'm choosing a surjective T.
Thus, AT=A'T implies A=A', hence those homomorphisms really are new.
Then, as far as I can see, the number of surjective homomorphisms from
G to G/K is equal to |aut(G/K)|.
So my claim that the number of homomorphisms from G to C is equal to
m*|G/K| where m=|aut(G/K|
still seems right.
quasi |
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Rupert
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| Posted: Mon May 28, 2007 11:01 am Post subject: Re: homomorphisms from a finite group to C* |
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On May 28, 3:15 pm, quasi <q...@null.set> wrote:
| Quote: |
On Mon, 28 May 2007 00:01:33 -0500, quasi <q...@null.set> wrote:
On 27 May 2007 20:57:11 -0700, Keith Ramsay <kram...@aol.com> wrote:
On May 27, 10:42 pm, quasi <q...@null.set> wrote:
|Finally, prove it assuming G is nonabelian. Let K be the commutator
|subgroup of G. Show that every homomorphism from G/K to C corresponds
|to k distinct homomorphisms from G to C where k=|K|.
No, it only is true if G is Abelian. If G is not Abelian,
there are only |G/K|.
Keith Ramsay
Of course, I should have seen that.
For example, if K=G, then the only homomorphism from G to C is the
trivial one.
But wait, I take it back.
I'm not so sure you're right.
Every homomorphism from G to C factors through G/K, but there may be
more than one homomorphism from G onto G/K.
In fact, start with any homomorphism of G onto G/K and follow it with
an arbitrary automorphism of G/K to get a new homomorphism from G onto
G/K.
So I think the correct answer for the nonabelian case is m*|G/K| where
m=|aut(G/K)|.
quasi- Hide quoted text -
- Show quoted text -
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A homomorphism from G to C* is constant on every coset in G/K. Thus
there is a bijection from Hom(G,C*) to Hom(G/K,C*). |
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