Inclusion Mapping on Subgroup is Homomorphism

Theorem

Let $\struct {G, \circ}$ be a group.

Let $\struct {H, \circ_{\restriction H} }$ be a subgroup of $G$.

Let $i: H \to G$ be the inclusion mapping from $H$ to $G$.

Then $i$ is a group homomorphism.

Proof

Let $x, y \in H$.

From Group Axiom $\text G 0$: Closure, $x \circ_{\restriction H} y \in H$.

Then:

\(\ds \map i {x \circ_{\restriction H} y}\)

\(=\)

\(\ds x \circ_{\restriction H} y\)

Definition of Inclusion Mapping

\(\ds \)

\(=\)

\(\ds x \circ y\)

Definition of Restriction

\(\ds \)

\(=\)

\(\ds \map i x \circ \map i y\)

Definition of Inclusion Mapping

Hence the result by definition of group homomorphism.

$\blacksquare$

Sources

• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{II}$: Groups: Morphisms