# If $\phi$ is an injective homomorphism $G \rightarrow G'$ is it true that $G \cong \phi (G)$

A professor of mine wrote this in our notes and it does not seem quite right to me. I was wondering if someone could say whether this is true or not.

$$\begin{array}{l}{\text { If } \phi : G \rightarrow G^{\prime} \text { is a homomorphism, then } \phi(G) \text { can be thought of as } G \text { 'partially }} \\ {\text { collapsed.' If } \phi \text { is an injection, then } G \cong \phi(G) .}\end{array}$$

It does not seem to me that just because $$\phi$$ is an injective homomorphism that $$G\cong \phi(G)$$. The reason I think this is because couldn't $$G$$ and $$\phi(G)$$ not have the same cardinality (ie. we need a bijection)?

Yes, by the first isomorphism theorem. $$(G/\operatorname{ker}\phi)\cong \phi(G)$$. But if $$\phi$$ is injective, $$\operatorname{ker}\phi=e$$.
• That makes sense. Though is $G/e = G$. It seems like the first is a set of singleton sets while the second is a set of elements? – Jac Frall Apr 1 at 2:10
• Yes. $G/e\cong G$. – Chris Custer Apr 1 at 2:13
Note that if $$\phi : G \to G^\prime$$ is injective, then $$\phi$$ is automatically a bijection $$G \to \phi(G)$$. Indeed, any function $$f : X \to Y$$ is by definition a surjection $$X \to f(X)$$. Since $$\phi$$ is also injective when understood as a map $$G \to \phi(G)$$, we see that $$G \cong \phi(G)$$.