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 \(uN = vN\).

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