Quotient Structure of Abelian Group is Abelian Group

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\mathcal R$ be a congruence relation on an abelian group $\left({G, \circ}\right)$.


Then the quotient structure $\left({G / \mathcal R, \circ_\mathcal R}\right)$ is an abelian group.


Proof

From Quotient Structure of Group is Group $\left({G / \mathcal R, \circ_\mathcal R}\right)$ is a group.

Let $\left[\!\left[{x}\right]\!\right]_\mathcal R, \left[\!\left[{y}\right]\!\right]_\mathcal R \in S / \mathcal R$.

\(\displaystyle \left[\!\left[{x}\right]\!\right]_\mathcal R \circ_{S / \mathcal R} \left[\!\left[{y}\right]\!\right]_\mathcal R\) \(=\) \(\displaystyle \left[\!\left[{x \circ y}\right]\!\right]_\mathcal R\) Definition of operation induced on $S / \mathcal R$ by $\circ$
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{y \circ x}\right]\!\right]_\mathcal R\) $\circ$ is commutative
\(\displaystyle \) \(=\) \(\displaystyle \left[\!\left[{y}\right]\!\right]_\mathcal R \circ_{S / \mathcal R} \left[\!\left[{x}\right]\!\right]_\mathcal R\) Definition of operation induced on $S / \mathcal R$ by $\circ$

Hence $\circ_{S / \mathcal R}$ is commutative.


Hence $\left({G / \mathcal R, \circ_\mathcal R}\right)$ is an abelian group.

$\blacksquare$


Sources