The Set Product Is A Binary Operation On Cosets Of Normal Subgroups Josh

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Solution

Let $G,N$ be groups such that $N\leq G$. Suppose $N\trianglelefteq G$. Let $a,b\in G$. We must show $(Na)(Nb)=N(ab)$. Let $x\in (Na)(Nb)$. Pick $n_{1},n_{2}\in N$ such that $x=(n_{1}a)(n_{2}b)$. We know that $x=(n_{1}a)(n_{2}b)=n_{1}(an_{2})b$ because associativity holds in $G$. Because $N\trianglelefteq G$ we know $Na=aN$. It follows by properties of cosets that $an_{2}\in Na$. Pick $n_{3}\in N$ such that $an_{2}=n_{3}a$. We now compute $$x=n_{1}(an_{2})b=n_{1}(n_{3}a)b=(n_{1}n_{3})(ab).$$ We know that $n_{1}n_{3}\in N$ and that $ab\in G$ because $N$ and $G$ are both closed under the operation. It follows that $x=(n_{1}n_{3})(ab)\in N(ab)$. Thus, $(Na)(Nb)\subseteq N(ab)$.

Now let $y\in N(ab)$ such that $y=(n_{4}n_{5})(ab)$ for $n_{4}=e_{N}$, the identity in $N$, and $n_{5}\in N$. Then we know that $y=(n_{4}n_{5})(ab)=n_{4}(n_{5}a)b$. We also know that $n_{5}a\in aN$. Pick $n_{6}\in N$ such that $an_{6}=n_{5}a$. Now we compute $$y=n_{4}(n_{5}a)b=n_{4}(an_{6})b=(n_{4}a)(n_{6}b)\in (Na)(Nb).$$ Therefore, we know $N(ab)\subseteq (Na)(Nb)$. Because $(Na)(Nb)\subseteq N(ab)$ and $N(ab)\subseteq (Na)(Nb)$ we conclude $(Na)(Nb)=N(ab). \square$

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