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Consider the following:

Let $A$ and $B$ be groups and $\phi:\ A\ \longrightarrow\ B$ a group homomorphism. Let $T$ be a normal subgroup of $\operatorname{im}\phi$, and let $Q = \lbrace x \in A :\ \phi(x) \in T \rbrace $.

I have managed to show that $Q$ is a normal subgroup of A. Now I conjecture:

$$ \frac{A}{Q} = \frac{\text{im}(\phi)}{T} $$

But I'm not sure how to show this in general.

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    $\begingroup$ In general, $T$ need not be a subgroup of $\operatorname{im}\phi$, so this cannot be true! $\endgroup$
    – Servaes
    Sep 23 '15 at 7:36
  • $\begingroup$ I re-read what I had done and I realized I didn't need it to be a normal subgroup of B, but just the image of $\phi$ $\endgroup$ Sep 23 '15 at 7:38
  • $\begingroup$ @Servaes thanks for pointing that out! $\endgroup$ Sep 23 '15 at 7:39
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Given the map $\phi$, by the first isomorphism theorem we have an isomorphism $$\overline{\phi}:\ A/\ker\phi\ \longrightarrow\ \operatorname{im}\phi.$$ Because $T$ is a normal subgroup of $\operatorname{im}\phi$ we have a surjective group homomorphism $$\pi:\ \operatorname{im}\phi\ \longrightarrow\ \operatorname{im}\phi/T.$$ Because $\phi$ and $\pi$ are both surjective, also $\psi:=\pi\circ\phi$ is a surjective group homomorphism $$\psi:\ A\ \longrightarrow\ \operatorname{im}\phi/T,$$ so $\operatorname{im}\psi=\operatorname{im}\phi/T$, and $\ker\psi=Q$ by definition. So again by the first isomorphism theorem we have a group isomorphism $$\overline{\psi}:\ A/Q\ \longrightarrow\ \operatorname{im}\phi/T.$$

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  • $\begingroup$ This is clever, I didn't think of reusing the first isomorphism theorem to prove its own generalization $\endgroup$ Sep 23 '15 at 7:48
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    $\begingroup$ I guess this shows that your generalisation is in fact equivalent to the first isomorphism theorem :) $\endgroup$
    – Servaes
    Sep 23 '15 at 7:49

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