The book I am reading has the proves the following theorem:
Let $G$ be a group and let $H$ be a normal subgroup of $G$. The set $G/H= (aH | a \in G)$ is a group under the operation $(aH)(bH) = abH$.
The first step they show is that: $$f: G/H \times G/H \rightarrow G/H \\ aH * bH \mapsto abH$$ is well defined. I tried this verification myself, but its a little different from the one in my book. I wanted to know if it was correct:
To check if $f$ is well defined, we consider the following two mappings: $$aH * bH \mapsto abH \\ a'H * b'H \mapsto a'b'H$$ If $aH=a'H$ and $bH=b'H$ then we want $abH=ab'H$.
So suppose $aH=a'H$ and $bH=b'H$. Then $aH * bH = a'H * b'H = a'b'H$. We also know $aH * bH=abH$. Thus when $aH=a'H$ and $bH=b'H$, it follows $abH=a'b'H$.
Is this correct?
This isn't correct. In the line "$\dots aH * bH = a'H * b'H \dots$" you explicitly assume that $f(aH,bH) = f(a'H,b'H)$, but this is what we want to prove. Another clue pointing out the mistake is that you never used the fact that $H$ is a normal subgroup of $G$, which is a must whenever $(aH)(bH) = (ab)H$ is well-defined.
To show well-definedness you have to prove that $abH = a'b'H$, without referring to the function $f$. The best way of doing this is to note that $a=a'h_1$, $b=b'h_2$. Now as $H \unlhd G$ $h_1b'=h_3b'$ and then:
$$ab = a'h_1b'h_2 = a'b'h_3h_2 = (a'b')h_3h_2 \implies ab \in (a'b')H \implies (ab)H = (a'b')H$$
or even shorter using cosets:
$$(ab)H = a(Hb) = (a'H)b = (a'b')H$$