Proposition 6

Let $G$ be a group with operators and $G'$ a magma with an action by $\Omega$, written exponentially. Let $f : G \to G'$ be a homomorphism of the magna $G$ into the magma $G'$ such that, for all $\alpha \in \Omega$ and all $x \in G$, $f(x^\alpha) = f(x)^\alpha$. Then $f(G)$ is a stable subset of $G'$ under the law on $G'$ and the action of $\Omega$; the set $f(G)$ with the induced laws is a group with operators and the mapping $x \mapsto f(x)$ of $G$ into $f(G)$ is a homomorphism of groups with operators.

By virtue of $§ 1$, no. 4, Proposition 1, $f(G)$ is a stable subset of $G'$ under the internal law on $G'$. For every element $x \in G$ and for every operator $\alpha$, $f(x)^\alpha = f(x^\alpha) \in f(G)$ and therefore $f(G)$ is stable under the action of $\Omega$ on $G'$. Writing the internal law of $G'$ multiplicatively,

for all elements $x, y, z$ in $G$; therefore the induced law on $f(G)$ is associative.

Let $e$ be the identity element of $G$. Its image $f(e)$ is an identity element of $f(G)$ ($§ 2$, no. 1). Every element of $f(G)$ is invertible in $f(G)$ ($§ 2$, no. 3). Therefore the law induced on $f(G)$ by the internal law on $G'$ is a group law. For all elements $x$ and $y$ in $G$ and every operator $\alpha$,

which shows that the action of $\Omega$ is distributive with respect to the group law on $f(G)$. Therefore $f(G)$ with the induced laws is a group with operators and clearly the mapping $x \mapsto f(x)$ is a homomorphism of groups with operators.