5

I'm trying to show that the quotient of the Heisenberg group with it's own center, H/Z(H), is abelian. I'm not entirely sure what makes up this quotient group in the first place though... and I'm a little confused as to what quotients of matrix groups with multiplicative operators look like. Help, thanks

ConfusedEngineer
  • 155

1 Answers1

5

Hints:

Show that

$$Z(H)=\left\{\;\begin{pmatrix}1&0&x\\0&1&0\\0&0&1\end{pmatrix}\;;\;x\in\Bbb F\;\right\}$$

with $\;\Bbb F=\;$ the field over which the Heisenberg Group is defined.

Now just check the homomorphism

$$\phi:H\to \Bbb F\times\Bbb F\;,\;\;\phi\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}:=(a,c)$$

Another way, when $\;\Bbb F=\Bbb F_p\;,\;\;p=$ a prime: we have that $\;H\;$ is a non-abelian group of order $\;p^3\;$ and thus $\;H'=Z(H)\;$ both by the Class Equation (i.e., a finite $\;p-$ group has a non-trivial center), and because $\;G/Z(G)\;$ cannot be cyclic non-trivial for any group $\;G\;$.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • How does showing this homomorphism prove that the quotient is abelian? – ConfusedEngineer Mar 25 '14 at 23:49
  • 1
    The homomorphism is surjective and $;\Bbb F\times \Bbb F;$ is an abelian group ... – DonAntonio Mar 26 '14 at 00:11
  • @user3749105 It is written there (almost 10 years ago!) that $;H'=Z(H);$ , so that the quotient is in fact $;H/H';$ and thus of course abelian. What part isn't clear to you? – DonAntonio Nov 19 '23 at 07:56
  • How does the fact that it was written here 10 years ago matter? Don't be arrogant! – user3749105 Nov 21 '23 at 20:05
  • @user3749105 First, good you erased the bad word you wrote before. Second, it is not about arrogance but about trying to remember what was done here. Third, now you understand how the quotient is abelian? – DonAntonio Nov 22 '23 at 15:41
  • @user3749105 Well, it was afortunate to you, otherwise you could easily have been banned from this site. – DonAntonio Nov 23 '23 at 13:39