Let $\phi:G \to G'$ be ANY group homomorphism.
If $H'$ is a subgroup of $G'$, then:
1) $H = \phi^{-1}(H')$ is a subgroup of $G$,
2) $\text{ker }\phi \subseteq H$.
To establish (1), let $a,b \in H$.
Since $\phi(\phi^{-1}(H')) = H'$, we have $\phi(a),\phi(b) \in H'$.
Since $H'$ is a subgroup of $G'$ we have $\phi(a)[\phi(b)]^{-1} \in H'$.
Since $\phi$ is a HOMOMORPHISM:
$\phi(a)[\phi(b)]^{-1} = \phi(a)\phi(b^{-1}) = \phi(ab^{-1})$.
Hence $ab^{-1} \in H$, since $\phi(ab^{-1}) \in H'$, and (1) is proven.
To see (2), suppose that $k \in \text{ker }(\phi)$, and let $e'$ denote the identity of $G'$.
Then $\phi(k) = e' \in H'$, whence $k \in \phi^{-1}(H') = H$, so $\text{ker }\phi \subseteq H$.
********************
Now, let $G' = G/M$, and $\phi = \gamma$. We have just two things left to show:
a) $\gamma^{-1}(H')$ is proper if $H'$ is a proper subgroup of $G/M$
b) $\gamma^{-1}(H')$ is normal if $H'$ is.
Let's do (b) first. Since $H'$ is normal, we have $x'H'x'^{-1} = H'$, for any such $x' \in G/M$.
Letting $H = \gamma^{-1}(H')$ (see above), we have, for any $g \in G, h\in H$:
$\gamma(ghg^{-1}) = \gamma(g)\gamma(h)[\gamma(g)]^{-1}$.
Now $\gamma(h) \in \gamma(\gamma^{-1}(H')) = H'$, and since $H'$ is normal, it follows that:
$\gamma(ghg^{-1}) \in H'$, so that $ghg^{-1} \in H$; that is, $H$ is normal.
Now, (a): let $x' \in G/M - H'$. Since $\gamma$ is surjective, $x' = gM$ for some $g \in G$.
I claim $g \not\in H$, for if so, then $\gamma(g) = gM = x' \in H'$, contradiction. Thus $H$ is proper.
********************
I note in passing that the WHOLE POINT of the Fundamental Isomorphism Theorem is that any homomorphism:
$\phi: G \to \phi(G)$ is of the form:
$\gamma: G \to G/M$ for some normal subgroup $M$ of $G$: namely, the kernel of $\phi$.
Normal subgroup = kernel
Quotient group = homomorphic image
The LHS of these two things are in "object language", the RHS are in "mapping language". Often the quotient group is regarded AS the projection mapping, and the kernel is regarded AS the inclusion mapping.
When we take the quotient homomorphism:
$\Bbb Z \to \Bbb Z_n = \Bbb Z/n\Bbb Z$, what we are doing is effectively "setting $n$ to 0".
When we write an integer as:
$a = nq + r$, it is apparent that:
$a + a' = nq + r + nq' + r' = n(q+q') + (r + r')$.
So the mapping that sends an integer $a$ to its (UNIQUE!) remainder upon division by $n$ is a homomorphism (it preserves addition).
This changes looking at an infinite set, to a finite set. There are only $n$ integers between one multiple of $n$, and the next (including either the first, or the last). Regarding these as COSETS:
$k + n\Bbb Z$
is cumbersome, it is far easier to regard them as "remainders".
Here is how one uses this, in practice:
Solve:
$7x + 11y = 100$ for $x,y \in \Bbb Z^+$ (this is a Diophantine equation).
Take mod 11:
$7x = 1$ (mod 11).
$\text{gcd}(7,11) = 1$, so there are integers $a,b$ with:
$7a + 11b = 1$. Use Euclidean division to find:
$11 = 1(7) + 4$
$7 = 1(4) + 3$
$4 = 1(3) + 1$, so that:
$1 = 4 - 3 = 4 - (7 - 4) = 2(4) - 7 = 2(11 - 7) - 7 = 2(11) - 3(7)$ so that $a = -3, b = 2$.
Since $-3 = 8$ (mod 11), we have $7^{-1} = 8$ (mod 11), thus:
$8(7x) = 8$ (mod 11)
$x = 8$ (mod 11).
So, in our original equation we have:
$7x + 11y = 100$
$7(8 + 11k) + 11y = 100$
$56 + 11(7k) + 11y = 99 + 1$
$55 + 11(7k) + 11y = 99$
$5 + 7k + y = 9$
$y = 4 - 7k$.
If $y > 0$, then $k \leq 0$.
If $x > 0$ then $k \geq 0$.
So the only $k$ that works is $k = 0$, and $(x,y) = (8,4)$.