Left (and Right) Cosets form a Partition in G

Theorem

Let $N$ be any Subgroup of the Group $G$ . The set of left cosets of $N$ in $G$ form a partition of $G$ . Furthermore, for all $u, v \in G$ then $u N = v N$ iff $v^{- 1} u \in N$ . In particular, $u N = v N$ iff $u, v$ are representatives of the same coset.

Proof

Note that since $N$ is a subgroup of $G$ then $1 \in N$ . Thus $g = g \cdot 1 \in g N$ for all $g \in G$ . Thus:

G = ⋃_{g \in G} g N

To show that distinct left cosets have empty intersection, assume $u N \cap v N \neq \emptyset$ . We show that $u N = v N$ . Let $x \in u N \cap v N$ . Then:

x = u n = v m

for some $n, m \in N$ . In the latter equality we can multiply both sides on the right by $n^{- 1}$ to get:

u = v m n^{- 1} = v m_{1} : m_{1} = m n^{- 1} \in N

Now for any element $u t$ of $u N$ (ie: $t \in N$ ) then:

u t = (v m_{1}) t = v (m_{1} t) \in v N

Thus $u N \subseteq v N$ . By interchanging $v, u$ here one similarly gets $v N \subseteq u N$ . Thus, $u N = v N$ . Thus, we have a contradiction since we assumed the cosets as being distinct. Thus their intersection must be empty.

For the other part of the proof, let $u, v \in G$ . Then:

\begin{aligned} u N = v N & ⟺ \exists n \in N (u = v n) \\ ⟺ v^{- 1} u \in N \end{aligned}

which is equivalent to saying that $v$ is a representative for $u N$ , thus $u N = v N$ .

☐