Proving a different version of the Third Isomorphism Theorem for groups Theorem: let $\phi_1:G_1\to G_2$ and $\phi_2:G_2\to G_3$ be surjective homomorphisms, then
$$\frac{G_1/\ker \phi_1}{\ker(\phi_2\circ\phi_1)/\ker(\phi_1)}\cong \frac{G_1}{\ker(\phi_2\circ\phi_1)}.$$

Proof idea:

*

*By the First Isomorphism Theorem there are isomorphisms
$$\psi_1:G_1/\ker \phi_1\to G_2:g_1\ker\phi_1\mapsto \phi_1(g_1)$$
$$\psi_2:G_2/\ker \phi_2\to G_3:g_2\ker\phi_2\mapsto \phi_2(g_2)$$
$$\psi:G_1/\ker (\phi_2\circ\phi_1)\to G_3:g_1\ker(\phi_2\circ\phi_1)\mapsto (\phi_2\circ\phi_1)(g_1).$$

*We have $\psi_1^{-1}(G_2)=G_1/\ker\phi_1$ and $\psi_1^{-1}(\ker\phi_2)=\ker(\phi_2\circ\phi_1)/\ker\phi_1$ so that
$$ \frac{G_2}{\ker\phi_2}\cong G_3\cong \frac{G_1}{\ker(\phi_2\circ\phi_1)} \Rightarrow \frac{G_1/\ker\phi_1}{\ker(\phi_2\circ\phi_1)/\ker\phi_1}\cong\frac{G_1}{\ker(\phi_2\circ\phi_1)}$$
as wanted.

$1)$ needs no demonstration. Regarding $2)$, I have been able to show that $\ker(\phi_1)\trianglelefteq \ker(\phi_2\circ\phi_1)$, but I have trouble showing that $\psi_1^{-1}(\ker\phi_2)=\ker(\phi_2\circ\phi_1)/\ker\phi_1$.

How may one complete the proof without using first the Third Isomorphism Theorem?

The theorem above follows naturally from the Third Isomorphism Theorem. Furthermore, it implies the Third Isomorphism Theorem: if we have normal subgroups $H,K$ of $G$ such that $K\leq H$, then taking $G_1=G$, $G_2=G/K$ and $G_3=G/H$ together with the maps $\phi_1:g\mapsto gK$ and $\phi_2:gK\mapsto gH$ one gets the Third Isomorphism Theorem as a corollary.
 A: Alternative (imo, better) answer: Since $\psi_1$ is an isomorphism, $\psi_1^{-1} (\ker \phi_2) = \ker \phi_2 \circ \phi_1 / \ker \phi_1$ is equivalent to $\ker \phi_2 = \psi_1(\ker \phi_2 \circ \phi_1 / \ker \phi_1)$ and for all $g_2 \in G_2,$ we can find $g_1 \in G_1$ st $g_2 = \phi(g_1)$, and thus\begin{align*} g_2 = \phi_1(g_1) \in \ker \phi_2 &\iff \phi_2(\phi_1(g_1)) = e_3 \\ &\iff g_1 \in \ker \phi_2 \circ \phi_1 \\&\iff g_1 \ker \phi_1 \in \ker \phi_2 \circ \phi_1 / \ker \phi_1 \\ &\iff g_2 = \phi_1(g_1) = \psi_1(g_1 \ker \phi_1) \in \psi_1(\ker \phi_2 \circ \phi_1 / \ker \phi_1).
\end{align*}
A: So we want to show $\psi_1^{-1} (\ker \phi_2 ) = \ker \phi_2 \circ \phi_1 / \ker \phi_1.$
$\subseteq$: Let $g_1 \ker \phi_1 \in \psi_1^{-1} (\ker \phi_2 ) \subseteq G_1 / \ker \phi_1.$ It suffices to show that $g_1 \in \ker \phi_2 \circ \phi_1.$ Since $g_1 \ker \phi_1 \in \psi_1^{-1} (\ker \phi_2 )$ we get $$\phi_1(g_1) = \psi_1 (g_1 \ker \phi_1) \in \ker \phi_2$$ and thus $\phi_2 ( \phi_1 (g_1)) = 1$.
$\supseteq$: Let $g_1 \ker \phi_1 \in \ker \phi_2 \circ \phi_1 / \ker \phi_1$. Now it suffices to show that $\psi_1 ( g_1 \ker \phi_1) \in \ker \phi_2.$ Since $\phi_1(g_1) = \psi_1(g_1 \ker \phi_1)$, this follows easily from the fact that $g_1 \in \ker \phi_2 \circ \phi_1$.
