# Generalized Isomorphism Theorem for Groups

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.

• In general, $T$ need not be a subgroup of $\operatorname{im}\phi$, so this cannot be true! Sep 23 '15 at 7:36
• 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 $\phi$ Sep 23 '15 at 7:38
• @Servaes thanks for pointing that out! Sep 23 '15 at 7:39

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 $$\overline{\psi}:\ A/Q\ \longrightarrow\ \operatorname{im}\phi/T.$$