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# Left cosets of N in G partition G

Let $$N \leq G$$.

Proposition. The left cosets of $$N$$ in $$G$$ partition $$G$$.

Proof: First we show their union is precisely $$G$$. Since $$N \leq G$$, $$1 \in N$$ so $$g \in gN$$. Hence

$\bigcup_{g \in G} gN = G.$

Next we show that distinct cosets have empty intersection. We argue by contraposition. Assume $$uN \cap vN \neq \emptyset$$. There exist $$m,n \in N$$ such that $$um = vn$$ . So $$m = u^{-1}vn \in N$$, and by closure we have $$u^{-1}v \in N$$ also.

Thus, for all $$x \in N$$, $$u^{-1}vx \in N$$. This implies that $$vx \in uN$$, so $$vN \subseteq uN$$.

A similar argument implies $$uN \subseteq vN$$. Therefore $$uN = vN$$.

Note by Jake Lee
5 months, 1 week ago