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
[Math] Show that the quotient of the Heisenberg Group with its center is abelian.
abstract-algebra
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Best Answer
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 $\;G\;$.