Identity of Affine Group of One Dimension

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 $\map {\mathrm {Af}_1} \R$.

$\blacksquare$