Group Epimorphism Preserves Generator

Theorem

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

Let $\phi: G \to H$ be a group epimorphism.

Let $A$ be a generator for $\struct {G, \circ}$.

Then $\phi \sqbrk A$ is a generator for $\struct {H, *}$.

Proof

By definition of generator:

$A$ is the intersection of all subgroups of $G$ containing $A$.

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Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.15$