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I am reading the proof of the Second Isomorphism Theorem on Dummit and Foote's Abstract Algebra. Could someone please explain how $\varphi$ is surjective?

If $(ab)B$ is any element of $AB/B$, I don't see what element of the domain $A$ is mapped to $(ab)B$.

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    $\begingroup$ Note: $(ab)B=aB=\phi(a)$. $\endgroup$ – quasi Aug 16 '18 at 18:13
  • $\begingroup$ I see. Thank you so much. Does this mean $AB/B=A/B$? $\endgroup$ – gladimetcampbells Aug 16 '18 at 18:16
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    $\begingroup$ $A/B$ is not meaningful unless $B$ is a subgroup of $A$. But note the conclusion of the stated theorem: $AB/B \cong A/(A\cap B)$. $\endgroup$ – quasi Aug 16 '18 at 18:25
  • $\begingroup$ That's right. Do you mean if $B$ is a subgroup of $A$, then the conclusion just boils down to $A/B\cong A/B$? $\endgroup$ – gladimetcampbells Aug 16 '18 at 19:15
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    $\begingroup$ Yes, so for that case, the theorem tells you nothing that you didn't already know. $\endgroup$ – quasi Aug 16 '18 at 19:18
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$(ab)B \sim aB$ since $xB \sim yB$ if there exists some $b \in B$ so that $x=yb$.

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