Cosets in Abelian Group
From ProofWiki
Theorem
Let $G$ be an abelian group.
Then every right coset modulo $H$ is a left coset modulo $H$.
That is:
- $\forall x \in G: x H = H x$
In an abelian group, therefore, we can talk about congruence modulo $H$ and not worry about whether it's left or right.
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\) | \(\displaystyle \forall x, y \in G: x^{-1} y = y x^{-1}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \left({x \ \equiv^l \ y \ \left({\bmod H}\right) \iff y \ \equiv^r \ x \ \left({\bmod H}\right)}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Congruence Modulo a Subgroup | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \left({x \ \equiv^l \ y \ \left({\bmod H}\right) \iff x \ \equiv^r \ y \ \left({\bmod H}\right)}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Congruence Modulo a Subgroup is an Equivalence, therefore Symmetric |
$\blacksquare$
Sources
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 6.1$: Example $113$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 37 \gamma$
- Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 42$