Induced Group

Theorem

Let $\left({T, \oplus}\right)$ be a group whose identity is $e_T$, and let $S$ be a set.

Let $\left({T^S, \oplus}\right)$ be the structure on $T^S$ induced by $\oplus$.

Then $\left({T^S, \oplus}\right)$ is a group, and the inverse of a given mapping $f$ is its Induced Structure Inverse $f^*$:

$\forall f \in T^S: \forall x \in S: f^* \left({x}\right) = \left({f \left({x}\right)}\right)^{-1}$

If $\left({T, \oplus}\right)$ is abelian, then so is $\left({T^S, \oplus}\right)$.

Proof

• From Induced Structure Associative and Induced Structure Identity, $\left({T^S, \oplus}\right)$ is associative and has an identity.

• $\left({T^S, \oplus}\right)$ is patently closed, because from the definition of the composition of mappings:

$\forall f, g \in T^S: f \oplus g \in T^S$

• We have that the inverse of $f$ is $f^*$ from Induced Structure Inverse.

• If $\left({T, \oplus}\right)$ is abelian, then $\oplus$ is commutative on $T$.

Then from Induced Structure Commutative, so is the operation it induces on $T^S$, and the result follows.

$\blacksquare$