Do we have to prove G is either cyclic or non-cyclic to find out whether it is Abelian?
I believe the examples on p178 help with this problem
You need only show a counterexample to the claim: "Given a normal Abelian subgroup $ \scriptstyle H $ of $ \scriptstyle G $ such that $ \scriptstyle G/H $ is Abelian, $ \scriptstyle G $ must also be Abelian." The book cites $ \scriptstyle D_3 $ as such a counterexample.
$ \scriptstyle D_3 $ has the following 6 elements:
$ \scriptstyle R_0\,\,:\,\,ABC\to ABC\,\,\,\,R_{120}\,\,:\,\,ABC\to CAB\,\,\,\,R_{240}\,\,:\,\,ABC\to BCA $
$ \scriptstyle F_1\,\,:\,\,ABC\to CBA\,\,\,\,F_2\,\,:\,\,ABC\to BAC\,\,\,\,F_3\,\,:\,\,ABC\to ACB $
Its Cayley table is:
| $ \textstyle D_3 $ | $ \textstyle R_0 $ | $ \textstyle R_{120} $ | $ \textstyle R_{240} $ | $ \textstyle F_1 $ | $ \textstyle F_2 $ | $ \textstyle F_3 $ | |
| $ \textstyle R_0 $ | $ \scriptstyle R_0 $ | $ \scriptstyle R_{120} $ | $ \scriptstyle R_{240} $ | $ \scriptstyle F_1 $ | $ \scriptstyle F_2 $ | $ \scriptstyle F_3 $ | |
| $ \textstyle R_{120} $ | $ \scriptstyle R_{120} $ | $ \scriptstyle R_{240} $ | $ \scriptstyle R_0 $ | $ \scriptstyle F_2 $ | $ \scriptstyle F_3 $ | $ \scriptstyle F_1 $ | |
| $ \textstyle R_{240} $ | $ \scriptstyle R_{240} $ | $ \scriptstyle R_0 $ | $ \scriptstyle R_{120} $ | $ \scriptstyle F_3 $ | $ \scriptstyle F_1 $ | $ \scriptstyle F_2 $ | |
| $ \textstyle F_1 $ | $ \scriptstyle F_1 $ | $ \scriptstyle F_3 $ | $ \scriptstyle F_2 $ | $ \scriptstyle R_0 $ | $ \scriptstyle R_{240} $ | $ \scriptstyle R_{120} $ | |
| $ \textstyle F_2 $ | $ \scriptstyle F_2 $ | $ \scriptstyle F_1 $ | $ \scriptstyle F_3 $ | $ \scriptstyle R_{120} $ | $ \scriptstyle R_0 $ | $ \scriptstyle R_{240} $ | |
| $ \textstyle F_3 $ | $ \scriptstyle F_3 $ | $ \scriptstyle F_2 $ | $ \scriptstyle F_1 $ | $ \scriptstyle R_{240} $ | $ \scriptstyle R_{120} $ | $ \scriptstyle R_0 $ |
Let $ \scriptstyle H = \{R_0,R_{120},R_{240}\} $. From the Cayley table of $ \scriptstyle D_3 $, it's clear that $ \scriptstyle H $ is Abelian. $ \scriptstyle|D_3:H|\,=\,2 $, so we know that $ \scriptstyle H\triangleleft D_3 $ (we proved this in the previous exercise). $ \scriptstyle D_3/H $ is then $ \scriptstyle \{H,F_1H\} $. The Cayley table for $ \scriptstyle D_3/H $ is:
| $ \scriptstyle D_3/H $ | $ \scriptstyle H $ | $ \scriptstyle F_1H $ | |
| $ \scriptstyle H $ | $ \scriptstyle H $ | $ \scriptstyle F_1H $ | |
| $ \scriptstyle F_1H $ | $ \scriptstyle F_1H $ | $ \scriptstyle H $ |
From this, it is obvious that $ \scriptstyle D_3/H $ is also Abelian. However, $ \scriptstyle D_3 $ is not Abelian. For example, $ \scriptstyle R_{120}F_1\,\,=\,\,F_2 $, but $ \scriptstyle F_1R_{120}\,\,=\,\,F_3 $. Thus, given a normal Abelian subgroup $ \scriptstyle H $ of $ \scriptstyle G $ such that $ \scriptstyle G/H $ is Abelian, $ \scriptstyle G $ need not be Abelian as well. $ \scriptstyle \Box $
- --Nick Rupley 03:04, 2 October 2008 (UTC)
