Is this textbook proof of the Third Homomorphism Theorem of groups correct? I'm reading Abstract Algebra, 3rd edition by Herstein. On page 87 (in section 2.7) he states The Third Homomorphism Theorem like this:

If the map $\phi:G \rightarrow G'$ is a homomorphism of $G$ onto $G'$ with kernel $K$ then, if $N' \lhd G'$ and $N = \{a \in G \mid \phi (a) \in N' \}$, we conclude that $G/N \simeq G'/N'$. Equivalently, $G/N \simeq (G/K)/(N/K)$.

He proves that $G/N \simeq G'/N'$ by defining the map $\psi:G \rightarrow G'/N'$ by $\psi(a) = N'\phi(a)$ and showing that it is a surjective homomorphism with kernel equal to $N$ and thus $G/N \simeq G'/N'$ by the First Homomorphism Theorem. This all is fine with me.
But as for the last statement of his theorem, he just says:

Finally, again by Theorems 2.7.1 [the First Homomorphism Theorem] and 2.7.2 [The Correspondence Theorem], $G' \simeq G/K$, $N' \simeq N/K$, which leads us to $G/N \simeq G'/N' \simeq (G/K)/(N/K)$.

I understand that $G' \simeq G/K$ and $N' \simeq N/K$, but I don't see how that implies $G'/N' \simeq (G/K)/(N/K)$. I tried to prove this general theorem: If $A \lhd G, B \lhd H, A \simeq B$, and $G \simeq H$, then $G/A \simeq H/B$. Then I realized it's not true. A simple counterexample is given by $G = H = \mathbb{Z}$, $A = \mathbb{Z}$, and $B = 2\mathbb{Z}$.  
Is Herstein mistakenly thinking that this statement to which I just provided a counterexample is a theorem? Or is there some other reason for his final conclusion?
By the way, I realized how to prove $G/N \simeq (G/K)/(N/K)$ directly without first proving $G/N \simeq G'/N'$ first. So, the validity of his original theorem is not an issue for me.
Thanks for any help you can give.
 A: The point is that the isomorphisms $\psi:G/K\to G'$ and $\psi_1:N/K\to N'$ are connected, because of the definition of $N$, namely $\psi_1$ is the restriction of $\psi$ to $N/K$. 
This additional condition allows you to prove the claim in itself. 
A: So the problem with your example is that the isomorphism from $\mathbb{Z} \rightarrow \mathbb{2Z} $ is not a restriction of the isomorphism from $\mathbb{Z} \rightarrow \mathbb{Z}$ if you pick it to be the identity map. So in general, if $G \simeq H$ (via $\psi: G \rightarrow H$), $A \simeq B,$ where $A, B$ are normal subgroups of $G, H$ respectively, and $\psi(A)= B$, then you have a commutative diagram:
$\require{AMScd}$
\begin{CD}
G @>{\psi}>> H\\
@VVV @VVV\\
G/A @>{\psi'}>> H/B
\end{CD}
where $\psi'$ is induced from $\psi$ by: $\psi'[g]= [\psi(g)]$, and if $[a] = [b]$ in $G/A$, then $ab^{-1} \in A$, which implies $\psi(ab^{-1})= \psi(a) \psi(b^{-1}) \in B$, which implies $\psi'$ is well defined and the diagram is commutative. Now, $\psi'$ is obviously onto since $\psi$ is an isomorphism and both quotient maps are onto. $\psi'$ is also injective since the kernel of $\psi'$ is {$[g]: \psi(g) \in B$}, which is the identity in $G/A$.
In your case, the restriction of the isomorphism  $G' \rightarrow G/K $ is $N' \rightarrow N/K$, so the claim is correct.
