First Isomorphism Theorem - hole in the proof from my book? I have a book that contains a proof for their so-called "First Isomorphism Theorem", which esentially states that if $f$ is a surjective group morphism $f:G\to G'$ and $H$ is a normal subgroup of $G$, then $G/H$ is isomorphic to $G'/f(H)$.
Their proof goes like this:
They consider $\pi: G' \to G'/f(H)$ to be the canonical map, $\pi(x) = xf(H)$, which is surjective. Then they take $f':G\to G'/f(H), \ f'=\pi \circ f$. Since $f$ and $\pi$ are surjective, $f'$ is also surjective. Thus, by a previously proven theorem in the book (called the 'Fundamental Isomorphism Theorem'), we have that $G/\ker f'$ is isomorphic to $G'/f(H)$.
Now they go on to prove that $\ker f' = H$ in order to finalize the proof.
The following sequence of equivalences holds:
$x \in \ker f' \Leftrightarrow \pi(f(x)) = 1 \Leftrightarrow f(x) \in \ker \pi = f(H)$. (the $1$ here is the unity for $G'/f(H)$, so $f(H)$).
So now we have $f(\ker f') =f(H)$ and now comes the part where I get stuck. They now call on a previously proven theorem (called the Correspondence Theorem for Normal Subgroups), which states that if we have a surjective group morphism $f:G\to G'$, then the map $H\to f(H)$ from the set of normal subgroups of $G$ that contain $\ker f$ to the set of normal subgroups of $G'$ is bijective), but this does not look correct as we have no knowledge of whether $H$ actually contains $\ker f$ ($\ker f'$ does).
I also thought of $\pi$'ing the equality and instead using $f'(\ker f') = f'(H)$, but we don't know that $\ker f'$ belongs to $H$ either).
**EDIT: ** Source of the faulty theorem: The Romanian book "Bazele Algebrei, vol. I" (English would probably be "Fundamentals of Algebra) by C. Năstăsecu, C. Niță and C. Vraciu, Bucharest, 1986. The page is 54 and the theorem I mentioned is Proposition 3.19.
 A: It looks like the authors forgot to add the assumption that $H$ contains $\ker (f)$ to the statement of the result.
One can find many cases where the assertion fails without an assumption like this. For example let $G$ be $\mathbb{Z}$ and let $H$ be the trivial subgroup. For any $n\geq 1$, we have a surjective homomorphism $f$ from $\mathbb{Z}$ to $G'=\mathbb{Z}/n\mathbb{Z}$, and $f(H)$ is still the trivial subgroup. In particular, $G/H$ and $G'/f(H)$ are not isomorphic.
Taking $n=1$ in the previous situation shows how drastic the error is. Indeed, given any nontrivial group $G$, consider the surjective homomorphism $f:G\to G'$, where $G'$ is trivial group. If $H$ is the trivial subgroup of $G$, then $G/H=G$ and $G'/f(H)$ is trivial.
More generally if this formulation of the First Isomorphism Theorem were true then it would imply that anytime there is a surjective homomorphism from $G$ to $G'$, then $G$ and $G'$ are isomorphic, which is absurd.
I might also suggest you edit your question to add the name of the book and page number. This could help future users who run into the same problem.
