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I have a question about the definition of the Isomorphism theorem given my notes.

The Isomorphism Theorem: Let $\phi : G \rightarrow G'$ be a group homomorphism with kernel $K$. Then $\phi$ induces an isomorpism $\bar{\phi}: G/K \rightarrow \phi (G')$ given by $\bar{\phi}(Kg) = \phi(g)$ for all $Kg$ in $G/K$ ...

However shouldn't the part be "for all $g\in G$"? Is this the same thing?

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  • $\begingroup$ Why you call it definition of the isomorphism theorem? It sounds a bit strange to me. You just give the theorem a name, but you don't define something new. I wouldn't call it a definition. $\endgroup$ – user60589 Nov 26 '15 at 16:16
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    $\begingroup$ The domain of $\overline{\phi}$ consists of cosets $Kg$, not elements $g$ of $G$. In fact, to prove $\overline{\phi}$ is well-defined, it is necessary to show that $\phi$ is constant on the entire coset $Kg$, and that it does not matter which representative $kg$ we use. $\endgroup$ – David Wheeler Nov 26 '15 at 16:32
  • $\begingroup$ @user60589 Statement of isomorphism theorem would be more correct, I guess. $\endgroup$ – M Turgeon Nov 26 '15 at 16:40
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Formally, the statement is more correct. If you define a function $f:A\longrightarrow B$, you must specify the action of $f$ over a generic element of $A$.

In your case, a more pompous statement (but formally more precise) should be: $\bar{\phi}:G/K\longrightarrow \phi(G)$ is defined in the following way: for every coset $\alpha \in G/K$, let be $g\in G$ a representative of $\alpha$, id est such that $\alpha=Kg$, let be $$\bar{\phi}(\alpha)=\bar{\phi}(Kg)=\phi(g)$$

The solution you provide is perfectly correct, since $G/K=\{Kg\mid g\in G\}$.

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It is the same thing because $Kg$ runs over $G/K$ when $g$ runs over $G$.

However, the key point is that $\bar{\phi}(Kg) = \phi(g)$ is well defined, that is, does not depend on the choice of the representative $g$ for the coset $Kg$. This follows because $K = \ker \phi$, of course.

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Do you understand the line " induces an isomorphism...." ?

Your statement of the 1st Isomorphism Theorem does not mention the canonical homomorphism G to G/K , nor the factorisation of the homomorphism G to G'.

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