Conjugate of a Set by Identity
From ProofWiki
Theorem
Let $\left({G, \circ}\right)$ be a group whose identity is $e$.
Let $S \subseteq G$.
Then the conjugate of $S$ by $e$ is $S$:
- $S^e = S$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle S^e\) | \(=\) | \(\displaystyle \left\{ {y \in G: \exists x \in S: y = e \circ x \circ e^{-1} }\right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Conjugate of a Set | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \left\{ {y \in G: \exists x \in S: y = x}\right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Identity | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
$\blacksquare$
Sources
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 45$