Is ${G}/\ker(\phi)$ uniquely isomorphic to $\phi(G)$?

Question

Let $\phi:G \rightarrow G'$ be a group homomorphism. Then, by the First Isomorphism Theorem for groups, $\frac{G}{\ker(\phi)}$ is isomorphic to $\phi(G)$. For ease of notation, denote $\ker(\phi)$ by $K$. Is the "usual" isomorphism, $\psi: \frac{G}{K}\rightarrow\phi(G)$ defined as $\psi(aK)=\phi(a)$ the only isomorphism?

Answer 1

While it is true that the isomorphism $\psi: G/ \ker(\phi) \to \text{im}(\phi)$ is not the only one (as mentioned above), the reason $\psi$ is often called "usual" or "canonical" is that it is the unique isomorphism (even homomorphism) that is compatible with the projection $p : G \to G / \ker(\phi)$ and inclusion $j : \text{im}(\phi) \to G'$.

Explicitly, given the homomorphism $\phi : G \to G'$, one can form the following commutative diagram.

$ \require{AMScd} \begin{CD} \ker(\phi) @>{i}>> G @>{\phi}>> G' \\ & & @V{p}VV & @AA{j}A \\ & & G / \ker{\phi} @>{\psi}>> \text{im}(\phi) \end{CD} $

If $f: G / \ker(\phi) \to \text{im}(\phi)$ is any another homomorphism making the diagram commute (i.e. such that $j \circ f \circ p = \phi$), because $p$ is surjective and $j$ is injective we can right-cancel the former and left-cancel the latter from the equation $j \circ f \circ p = j \circ \psi \circ p$ to otain that $f = \psi$.

Answer 2

As mentioned in the comments above, one can simply (right-)compose $\psi$ with an automorphism of $\frac{G}{K}$ (call this $\pi$) and obtain an isomorphism, $\psi \circ \pi: \frac{G}{K} \rightarrow G'$. Every group with more than two elements will have a non-trivial automorphism, and we can use that as $\pi$, thereby obtaining a "new" isomorphism $\psi\circ\pi$. For the groups of orders one and two, the only automorphism is the trivial/identity automorphism. In such a case, $\psi$ is the only isomorphism.