How can you tell whether two groups are homomorphic/isomorphic?

Question

Suppose you have two groups, $G$ and $H$. I've been taught the following definitions:

"$G$ is homomorphic to $H$ iff there exists some function $\theta$ which gives the mapping $\theta : G \rightarrow H$

$G$ is isomorphic for $H$ iff, in addition to this, $\theta : G \rightarrow H$ is a bijection"

How are you supposed to know whether the function $\theta$ exists, or what it is?

Answer 1

If two groups with binary operation $(G, \cdot)$ and $(H, \star)$ are homomorphic then there $\textit{exists}$ a mapping $\theta:G\rightarrow H$. More precisely, $G$ and $H$ are homomorphic if there exists a mapping $\theta$ such that

$\theta(a \cdot b) = \theta(a)\star \theta(b)$ for any $a,b \in G$.

Often we don't know what this mapping is, but we find one and show it satisfies this property (i.e., preservation of the group operation), to show that the groups are homomorphic, i.e., there exists a homomorphism from one group to the other. An isomorphism is just a bijective homomorphism (both injective and surjective). If two groups are isomorphic then for all intents and purposes they can be considered the same group, because they have the same group structure. The elements are just named differently.

Answer 2

It is not easy, in general, to tell whether two groups are isomorphic. It is, however, quite easy to show that any two groups are "homomorphic."

Let $G, H$ be groups. Then I will define $\theta: G \to H$ by $\theta(g) = e_H$ for all $g \in G$, where $e_H$ is the identity element of $H$. Then for any $g, g' \in G$, $\theta(g g') = e_H = e_H e_H = \theta(g) \theta(g')$, so $\theta$ is indeed a group homomorphism.