Product of normal cosets is well-defined
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Let $N$ be a normal subgroup of $G$. This means that $gN = Ng$ for all $g \in G$. The coset of a normal subgroup is called a normal coset.
Define the product of cosets $aN$ and $bN$ as \[ (aN)(bN) = (ab)N \]
This product is well-defined, which means that it does not depend on the representation of the coset. This means that \[ a_1N = a_2N \wedge b_1N = b_2N \Rightarrow (a_1N)(b_1N) = (a_2N)(b_2N) \]
Proof
\begin{align} & (a_1N)(b_1N) &= (a_1b_1)N \\ &= a_1(b_1N) &= a_1(b_2N) \\ &= a_1(Nb_2) &= (a_1N)b_2 \\ &= (a_2N)b_2 &= a_2(Nb_2) \\ &= a_2(b_2N) &= (a_2b_2)N \\ &= (a_2N)(b_2N) \end{align}
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- Number of transitive dependencies: 5