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?