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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)?

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Yes, by the first isomorphism theorem. $(G/\operatorname{ker}\phi)\cong \phi(G)$. But if $\phi$ is injective, $\operatorname{ker}\phi=e$.

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  • $\begingroup$ 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? $\endgroup$ – Jac Frall Apr 1 at 2:10
  • $\begingroup$ Yes. $G/e\cong G$. $\endgroup$ – Chris Custer Apr 1 at 2:13
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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 $G \cong \phi(G)$.

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