First Isomorphism Theorem: Does each homomorphism has to be surjective? Is it possible to define an homomorphism $\phi:G\to H$ such that $|G|<|H|$?

Question

I have a little bit of misunderstanding about homomorphism and the first isomorphism.

Does each homomorphism has to be surjective? Is it possible to define an homomorphism $\phi:G\to H$ such that $|G|<|H|$?

If this is possible how am I supposed to use the the first isomorphism, since for each $N \triangleleft G $ we get $\frac{|G|}{|N|}<|H|$?

What's about the inifnity case, is it possible to define an homomorphism $\phi: G \to H$ such that $G<H$?

Thanks!

Answer 1

Homomorphisms can be injective, surjective, isomorphic, or neither of them. It only has to be invariant on the structure: $\varphi(a\cdot b)=\varphi(a)\cdot \varphi(b)$ in case of (multiplicatively written) groups, or continuous in case of topological spaces.

An instance for your example $|G|<|H| $ are subgroups $G\subseteq H,$ even if they are usually noted the other way around. If we have a subgroup, then $\iota :G\longrightarrow H$ with the embedding $\iota(g)=g\in H$ is an injective homomorphism (=monomorphism) that is not surjective in case of $G\subsetneq H$ is a proper subgroup, $G=\{1\}$ for example.

Answer 2

Another standard homomorphism which is not surjective (for $|G|\ge 3$), is Cayley's one $a\mapsto(g\mapsto ag)$ from $G$ into $S_G$.