I was trying to prove the First Isomorphism Theorem in an "elementary" way and this is what I done:
Let $G$ and $G'$ be two groups and $f:G \to G'$ be a homomorphism. Then $\ker f$ is a normal subgroup of $G$ and $G/\ker f \cong \mathrm{Im} f$.
Proof. It's easy to prove that $\ker f$ and $\mathrm{Im} f$ are groups. In order to prove that $\ker f$ is a normal subgroup of $G$ we must prove that $x (\ker f) x^{-1}=\ker f$ for all $x \in G$. This is easy: Let $x \in G$ be arbitrarily chosen. For every $a \in \ker f$ we have $f(xax^{-1})=f(x)f(a)f(x^{-1})=f(x) \cdot 1' \cdot f(x^{-1})=f(x)f(x^{-1})=f(1)=1'$. Thus $x(\ker f)x^{-1} \subset \ker f$. Now, for every $b \in \ker f$ we have $b=x(x^{-1}bx)x^{-1} \in x (\ker f)x^{-1}$ (because we know that $x^{-1}bx \in \ker f$). So $\ker f \subset x(\ker f)x^{-1}$. Therefore $x(\ker f)x^{-1}= \ker f,~\forall~x \in G,$ so $\ker f$ is normal.
Now for every $g' \in \mathrm{Im}f$ let's define $A_{g'}=\{x \in G \mid f(g)=g'\}$. It's clear that these sets are pairwise disjoint and $\bigcup\limits_{g \in \mathrm{Im}f}A_{g'}=G$. Let $M= \{A_{g'} \mid g' \in G'\}$. Let's consider the structure $(M, \cdot)$ where the operation is defined as follows: $$A_x \cdot A_y = A_{xy}$$
It is not hard to prove that $(M, \cdot)$ is a group. Now let's consider the function $g:M \to \mathrm{Im}f$ as follows: $$g(A_x)=x,~ \forall~ x \in \mathrm{Im}f.$$
This function is well defined and we can easily see that $g$ is an isomorphism and the conclusion follows.
I'm not really sure if my approach is totally correct. I don't really see where the fact that $\ker f$ is normal was used in my proof. I know that it is essential in order for $G / \ker f$ to make sense. It might be related to the construction of the sets $A_x$ but I can't see it. I also used the word "easy" quite a lot in my proof so there is a chance that I missed something subtle.
I would appreciate if someone confirms the fact that my proof is correct, or, otherwise, if someone point out the mistakes.
Best Answer
You can have $A_x = A_{x'}$ and $A_y = A_{y'}$ for $x \neq x'$ and $y \neq y'$. Then your definition of the product gives $A_x A_y = A_{xy}$ but also $A_xA_y = A_{x'} A_{y'} = A_{x'y'}$ so the question becomes: Do we have $A_{xy}= A_{x'y'}$?
The answer to this question is: this holds if and only if our subgroup is normal. I leave the verification of this fact to you as a good exercise. Basically, normality is necessary to ensure that the product you defined makes sense.