Quotient of Group by Center Cyclic implies Abelian

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Theorem

Let $G$ be a group.

Let $\map Z G$ be the center of $G$.

Let $G / \map Z G$ be the quotient group of $G$ by $\map Z G$.

Let $G / \map Z G$ be cyclic.


Then $G$ is abelian, so $G = \map Z G$.

That is, the group $G / \map Z G$ cannot be a cyclic group which is non-trivial.


Proof

Suppose $G / \map Z G$ is cyclic.

Then by definition:

$\exists \tau \in G / \map Z G: G / \map Z G = \gen \tau$

Since $\tau$ is a coset by $\map Z G$:

$\exists t \in G: \tau = t \map Z G$

Thus each coset of $\map Z G$ in $G$ is equal to $\paren {t \map Z G}^i = t^i \map Z G$ for some $i \in \Z$.


Now let $x, y \in G$.

Suppose $x \in t^m \map Z G, y \in t^n \map Z G$.

Then $x = t^m z_1, y = t^n z_2$ for some $z_1, z_2 \in \map Z G$.

Thus:

\(\displaystyle x y\) \(=\) \(\displaystyle t^m z_1 t^n z_2\)
\(\displaystyle \) \(=\) \(\displaystyle t^m t^n z_1 z_2\) $z_1$ commutes with all $t \in G$ since it is in the center
\(\displaystyle \) \(=\) \(\displaystyle t^{m + n} z_1 z_2\)
\(\displaystyle \) \(=\) \(\displaystyle t^{n + m} z_2 z_1\)
\(\displaystyle \) \(=\) \(\displaystyle t^n t^m z_2 z_1\)
\(\displaystyle \) \(=\) \(\displaystyle t^n z_2 t^m z_1\) $z_2$ commutes with all $t \in G$ since it is in the center
\(\displaystyle \) \(=\) \(\displaystyle y x\)


This holds for all $x, y \in G$, and thus $G$ is abelian.

Thus by Group equals Center iff Abelian $\map Z G = G$.

Therefore Quotient of Group by Itself it follows that $G / \map Z G$ is the trivial group.

$\blacksquare$


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