Quotient Group of Cyclic Group/Proof 2

Theorem

Let $G$ be a cyclic group which is generated by $g$.

Let $H$ be a subgroup of $G$.


Then $g H$ generates $G / H$.


Proof

Let $H$ be a subgroup of the cyclic group $G = \gen g$.

Then by Homomorphism of Powers for Integers:

$\forall n \in \Z: \map {q_H} {g^n} = \paren {\map {q_H} g}^n = \paren {g H}^n$


As $G = \set {g^n: n \in \Z}$, we conclude that:

$G / H = q_H \sqbrk G = \set {\paren {g H}^n: n \in \Z}$


Thus, by Epimorphism from Integers to Cyclic Group, $g H$ generates $G / H$.

$\blacksquare$


Sources

  • 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.3$
This article is issued from ProofWiki. The text is available under Creative Commons Attribution-ShareAlike License unless otherwise noted. Additional terms may apply for the media files.