Identity of Affine Group of One Dimension
Jump to navigation
Jump to search
Theorem
Let $\map {\mathrm {Af}_1} \R$ denote the $1$-dimensional affine group on $\R$.
Then $\map {\mathrm {Af}_1} \R$ has $f_{1, 0}$ as an identity element.
Proof
Let $f_{a b} \in \map {\mathrm {Af}_1} \R$.
Then:
\(\ds \map {\paren {f_{a b} \circ f_{1, 0} } } x\) | \(=\) | \(\ds a \paren {1 x + 0} + b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a x + b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds f_{a b}\) |
\(\ds \map {\paren {f_{1, 0} \circ f_{a b} } } x\) | \(=\) | \(\ds 1 \paren {a x + b} + 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a x + b\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds f_{a b}\) |
Thus $f_{1, 0}$ is the identity element of $\map {\mathrm {Af}_1} \R$.
$\blacksquare$