order preserving group homomorphism

say you have two subgroups $\displaystyle G\text{ and }H$ and an order preserving group homomorphism $\displaystyle \eta:G\rightarrow H$ which maps the identity of $\displaystyle G$ to the identity of $\displaystyle H$

Can we show that $\displaystyle \eta(g)=g$ for all $\displaystyle g\in G$ so that we can identify $\displaystyle G$ as a subgroup of $\displaystyle H$ and $\displaystyle \eta$ is then just the inclusion?

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Originally Posted by Mauritzvdworm

say you have two subgroups $\displaystyle G\text{ and }H$ and an order preserving group homomorphism $\displaystyle \eta:G\rightarrow H$ which maps the identity of $\displaystyle G$ to the identity of $\displaystyle H$

Can we show that $\displaystyle \eta(g)=g$ for all $\displaystyle g\in G$ so that we can identify $\displaystyle G$ as a subgroup of $\displaystyle H$ and $\displaystyle \eta$ is then just the inclusion?

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The condition $\displaystyle \eta(g)=g\,,\,\forall g\in G$ is very strong, and it'd mean $\displaystyle G\subset H$ (i.e. not only as identification but as actual subset of a set...).

But from the given data yes: $\displaystyle \eta(g)=1\Longleftrightarrow g=1\Longleftrightarrow \eta\,\,is\,\,1-1$ and thus G is embedded in H.

Tonio