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

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?

• You can compose $\psi$ with an automorphism of $G/K$. Sep 17, 2020 at 17:39
• Right, of course...maybe a better question would be if the only automorphism would be the identity map. I'll edit. Sep 17, 2020 at 17:40
• The only automorphism need not be the identity. Counterexamples abound. Sep 17, 2020 at 17:41
• And every group with order more than two has a nontrivial automorphism? So that means we can have more than one isomorphism in all cases except when G itself is the trivial group or the group of order 2. Sep 17, 2020 at 17:48

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 $$f = \psi$$.
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.