Is the following a theorem? Let G be a group with center Z. If G/Z is abelian, then G is abelian. -- Gene
no - the Heisenberg group is a counterexample. It is nonabelian of order p^3; its center is of order p; and the quotient, of order p^2, is abelian. Cris On Nov 28, 2013, at 9:48 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Is the following a theorem?
Let G be a group with center Z. If G/Z is abelian, then G is abelian.
-- Gene _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The smallest examples are the quaternion group Q8 and the dihedral group D_8 of order 8. See http://en.wikipedia.org/wiki/Quaternion_group. The center of Q8 is {1,-1} and the quotient has order 4 so must be abelian. On Thu, Nov 28, 2013 at 11:48 PM, Eugene Salamin <gene_salamin@yahoo.com>wrote:
Is the following a theorem?
Let G be a group with center Z. If G/Z is abelian, then G is abelian.
-- Gene _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Yes, indeed the quaternions are the Heisenberg group where p=2 :-) On Nov 28, 2013, at 10:09 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
The smallest examples are the quaternion group Q8 and the dihedral group D_8 of order 8.
See http://en.wikipedia.org/wiki/Quaternion_group. The center of Q8 is {1,-1} and the quotient has order 4 so must be abelian.
On Thu, Nov 28, 2013 at 11:48 PM, Eugene Salamin <gene_salamin@yahoo.com>wrote:
Is the following a theorem?
Let G be a group with center Z. If G/Z is abelian, then G is abelian.
-- Gene _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
Thanks people. Confronted with this example, I was able to locate the mistake in my attempt at a proof . -- Gene
________________________________ From: Cris Moore <moore@santafe.edu> To: wclark@mail.usf.edu; math-fun <math-fun@mailman.xmission.com> Cc: Eugene Salamin <gene_salamin@yahoo.com> Sent: Thursday, November 28, 2013 9:11 PM Subject: Re: [math-fun] Group theory question
Yes, indeed the quaternions are the Heisenberg group where p=2 :-)
On Nov 28, 2013, at 10:09 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
The smallest examples are the quaternion group Q8 and the dihedral group
D_8 of order 8.
See http://en.wikipedia.org/wiki/Quaternion_group. The center of Q8 is {1,-1} and the quotient has order 4 so must be abelian.
On Thu, Nov 28, 2013 at 11:48 PM, Eugene Salamin <gene_salamin@yahoo.com>wrote:
Is the following a theorem?
Let G be a group with center Z. If G/Z is abelian, then G is abelian.
-- Gene _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________
math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
W. Edwin Clark pointed out to me that the p=2 Heisenberg group is the dihedral group of over 8, not the quaternions. Sorry for the mistake. Cris On Nov 29, 2013, at 10:13 AM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
Thanks people. Confronted with this example, I was able to locate the mistake in my attempt at a proof .
-- Gene
What started me on this path is the easy seen theorem that if G/Z is cyclic, then G is abelian. But then Z = G and G/Z is trivial. But note that G/Z is isomorphic to Inn(G) the inner isomorphism group of G. So no inner isomorphism group can be nontrivial cyclic, a nice fact. -- Gene
________________________________ From: Eugene Salamin <gene_salamin@yahoo.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, November 29, 2013 9:13 AM Subject: Re: [math-fun] Group theory question
Thanks people. Confronted with this example, I was able to locate the mistake in my attempt at a proof .
-- Gene
________________________________ From: Cris Moore <moore@santafe.edu> To: wclark@mail.usf.edu; math-fun <math-fun@mailman.xmission.com> Cc: Eugene Salamin <gene_salamin@yahoo.com> Sent: Thursday, November 28, 2013 9:11 PM Subject: Re: [math-fun] Group theory question
Yes, indeed the quaternions are the Heisenberg group where p=2 :-)
On Nov 28, 2013, at 10:09 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
The smallest examples are the quaternion group Q8 and the dihedral group
D_8 of order 8.
See http://en.wikipedia.org/wiki/Quaternion_group. The center of Q8 is {1,-1} and the quotient has order 4 so must be abelian.
On Thu, Nov 28, 2013 at 11:48 PM, Eugene Salamin <gene_salamin@yahoo.com>wrote:
Is the following a theorem?
Let G be a group with center Z. If G/Z is abelian, then G is abelian.
-- Gene
Gene, Such a group may not be abelian, but it *is* meta-abelian: http://en.wikipedia.org/wiki/Metabelian_group Victor On Sat, Nov 30, 2013 at 2:25 PM, Eugene Salamin <gene_salamin@yahoo.com>wrote:
What started me on this path is the easy seen theorem that if G/Z is cyclic, then G is abelian. But then Z = G and G/Z is trivial. But note that G/Z is isomorphic to Inn(G) the inner isomorphism group of G. So no inner isomorphism group can be nontrivial cyclic, a nice fact.
-- Gene
________________________________ From: Eugene Salamin <gene_salamin@yahoo.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, November 29, 2013 9:13 AM Subject: Re: [math-fun] Group theory question
Thanks people. Confronted with this example, I was able to locate the mistake in my attempt at a proof .
-- Gene
________________________________ From: Cris Moore <moore@santafe.edu> To: wclark@mail.usf.edu; math-fun <math-fun@mailman.xmission.com> Cc: Eugene Salamin <gene_salamin@yahoo.com> Sent: Thursday, November 28, 2013 9:11 PM Subject: Re: [math-fun] Group theory question
Yes, indeed the quaternions are the Heisenberg group where p=2 :-)
On Nov 28, 2013, at 10:09 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
The smallest examples are the quaternion group Q8 and the dihedral group
D_8 of order 8.
See http://en.wikipedia.org/wiki/Quaternion_group. The center of Q8 is {1,-1} and the quotient has order 4 so must be abelian.
On Thu, Nov 28, 2013 at 11:48 PM, Eugene Salamin < gene_salamin@yahoo.com>wrote:
Is the following a theorem?
Let G be a group with center Z. If G/Z is abelian, then G is abelian.
-- Gene
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
It is abelian. Every element of G can be written uniquely as (a_i z) where a_i ranges over a set of coset representatives, and z ranges over the center. But since G/Z is cyclic, a_i can be chosen as a^i, and then it is clear that all these elements of G commute. This theorem is exercise 3.3 in Rotman's Group Theory, 4th ed. -- Gene
________________________________ From: Victor Miller <victorsmiller@gmail.com> To: Eugene Salamin <gene_salamin@yahoo.com>; math-fun <math-fun@mailman.xmission.com> Sent: Saturday, November 30, 2013 11:39 AM Subject: Re: [math-fun] Group theory question
Gene, Such a group may not be abelian, but it *is* meta-abelian: http://en.wikipedia.org/wiki/Metabelian_group
Victor
On Sat, Nov 30, 2013 at 2:25 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
What started me on this path is the easy seen theorem that if G/Z is cyclic, then G is abelian. But then Z = G and G/Z is trivial. But note that G/Z is isomorphic to Inn(G) the inner isomorphism group of G. So no inner isomorphism group can be nontrivial cyclic, a nice fact.
-- Gene
________________________________ From: Eugene Salamin <gene_salamin@yahoo.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, November 29, 2013 9:13 AM
Subject: Re: [math-fun] Group theory question
Thanks people. Confronted with this example, I was able to locate the mistake in my attempt at a proof .
-- Gene
________________________________ From: Cris Moore <moore@santafe.edu> To: wclark@mail.usf.edu; math-fun <math-fun@mailman.xmission.com> Cc: Eugene Salamin <gene_salamin@yahoo.com> Sent: Thursday, November 28, 2013 9:11 PM Subject: Re: [math-fun] Group theory question
Yes, indeed the quaternions are the Heisenberg group where p=2 :-) It is abelian. Every element of G can be written uniquely as (a_i z)
where a_i ranges over a set of coset representatives, and z ranges over the center. But since G/Z is cyclic, a_i can be chosen as a^i, and then it is clear that all these elements of G commute. This theorem is exercise 3.3 in Rotman's Group Theory, 4th ed.
-- Gene
On Nov 28, 2013, at 10:09 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
The smallest examples are the quaternion group Q8 and the dihedral group
D_8 of order 8.
See http://en.wikipedia.org/wiki/Quaternion_group. The center of Q8 is {1,-1} and the quotient has order 4 so must be abelian.
On Thu, Nov 28, 2013 at 11:48 PM, Eugene Salamin <gene_salamin@yahoo.com>wrote:
Is the following a theorem?
Let G be a group with center Z. If G/Z is abelian, then G is abelian.
-- Gene
The question of what groups G can have a given center Z, i.e. what the "central extensions" of Z are, leads to a beautiful kind of group cohomology. To get a sense of it, choose a set of coset representatives T, with one in each coset of Z. Then write an element of g as g=hz, where z is in Z and h is in T. Now, when we multiply g=hz and g'=h'z', we get gg' = hzh'z' = hh' zz' Now hh' isn't generally in T. But it differs from an element of T by an element of Z, which is a function only of h and h'. Write hh' = h'' + f(h,h'). where h'' is in T. Thus gg' = h'' z'' where z'' = f(h,h') zz' Since Z is abelian, we can abuse notation and write z'' = z + z' + f(h,h'). So what's going on is that when we multiply two elements of G, their "central part" gets shifted by a function f, which depends on the cosets to which g and g' belong. Now for the fun part. Write down the axiom of associativity: that is, (g_1 g_2) g_3 = g_1 (g_2 g_3). When you work this out, you find that f(h,h') must obey a certain axiom as well. In terms of algebraic topology, f has to be a certain kind of "cocycle", and the axiom states that its "coboundary" is zero. To play with this, you can try to work out the possible central extensions of Z_2 by Z_2: that is, groups G that have a normal subgroup Z such that G/Z = Z = Z_2, and Z commutes with everything. Each one is characterized by a function f: Z_2 x Z_2 -> Z_2 which satisfies the axiom. This function is not unique, but there are two inequivalent cases: in one we get G = Z_2 x Z_2, and in the other we get Z_4. In fact, for Z_4 we can think of f as the carry bit when we add two integers mod 4! - Cris On Nov 30, 2013, at 12:25 PM, Eugene Salamin <gene_salamin@yahoo.com> wrote:
What started me on this path is the easy seen theorem that if G/Z is cyclic, then G is abelian. But then Z = G and G/Z is trivial. But note that G/Z is isomorphic to Inn(G) the inner isomorphism group of G. So no inner isomorphism group can be nontrivial cyclic, a nice fact.
-- Gene
________________________________ From: Eugene Salamin <gene_salamin@yahoo.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, November 29, 2013 9:13 AM Subject: Re: [math-fun] Group theory question
Thanks people. Confronted with this example, I was able to locate the mistake in my attempt at a proof .
-- Gene
________________________________ From: Cris Moore <moore@santafe.edu> To: wclark@mail.usf.edu; math-fun <math-fun@mailman.xmission.com> Cc: Eugene Salamin <gene_salamin@yahoo.com> Sent: Thursday, November 28, 2013 9:11 PM Subject: Re: [math-fun] Group theory question
Yes, indeed the quaternions are the Heisenberg group where p=2 :-)
On Nov 28, 2013, at 10:09 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
The smallest examples are the quaternion group Q8 and the dihedral group
D_8 of order 8.
See http://en.wikipedia.org/wiki/Quaternion_group. The center of Q8 is {1,-1} and the quotient has order 4 so must be abelian.
On Thu, Nov 28, 2013 at 11:48 PM, Eugene Salamin <gene_salamin@yahoo.com>wrote:
Is the following a theorem?
Let G be a group with center Z. If G/Z is abelian, then G is abelian.
-- Gene
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
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Cris Moore -
Eugene Salamin -
Victor Miller -
W. Edwin Clark