Quotient of Group by Center Cyclic implies Abelian
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:
\(\ds x y\) | \(=\) | \(\ds \paren {t^m z_1} \paren {t^n z_2 }\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds t^m \paren {z_1 t^n} z_2\) | Subset Product within Semigroup is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds t^m \paren {t^n z_1} z_2\) | $z_1$ commutes with all $t \in G$ since it is in the center | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {t^m t^n} \paren {z_1 z_2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds t^{m + n} \paren {z_1 z_2}\) | Powers of Group Elements: Sum of Indices | |||||||||||
\(\ds \) | \(=\) | \(\ds t^{n + m} \paren {z_2 z_1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {t^n t^m} \paren {z_2 z_1}\) | Powers of Group Elements: Sum of Indices | |||||||||||
\(\ds \) | \(=\) | \(\ds t^n \paren {t^m z_2} z_1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds t^n \paren {z_2 t^m} z_1\) | $z_2$ commutes with all $t \in G$ since it is in the center | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {t^n z_2 } \paren {t^m z_1 }\) | Subset Product within Semigroup is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds y x\) | by hypothesis |
This holds for all $x, y \in G$, and thus $G$ is abelian.
Thus by definition of abelian group:
- $\map Z G = G$
Therefore Quotient of Group by Itself it follows that $G / \map Z G$ is the trivial group.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 47 \epsilon$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 51.1$ The quotient group $G / Z$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Proposition $10.21$