Lemmas and Closure of Cosets

Lemma

For each $g \in G$ we have $g \in g H$ (see Left and Right Cosets)
For a given $g_{1}, g_{2} \in G$ we have $g_{1} \in g_{2} H ⟺ g_{2}^{- 1} g_{1} \in H$ .
For any pair of elements $g_{1}, g_{2} \in G$ we have either $g_{1} H \cap g_{2} H = \emptyset$ or $g_{1} H = g_{2} H$ . This means that the left-cosets Partition $G$

Proof

(1): $g = g \cdot 1 \in g H$ since $1 \in H$ . Thus $g \in g H$

(2):
$(\Rightarrow)$ : Suppose $g_{1} \in g_{2} H$ . Then:

g_{1} = g_{2} h

for some $h \in H$ . Now then:

\Rightarrow g_{2}^{- 1} g_{1} = h \in H

$(\Leftarrow)$ : Suppose $g_{2}^{- 1} g \in H$ . Then $g_{2}^{- 1} g_{1} = h$ for some $h \in H$ . Then:

g_{1} = g_{2} h \in g H

(3): If the sets are distinct then we are done, so suppose $g_{1} H \cap g_{2} H \neq \emptyset$ . We want to show that $g_{1} H = g_{2} H$ as a result. Choose some element:

g_{0} \in (g_{1} H) \cap (g_{2} H)

Let's show that $g_{1} H \subseteq g_{2} H$ . Take any $g_{1} h \in g_{1} H$ where $h \in H$ . Again, we want to show that $g_{1} h \in g_{2} H$ . Using (2), we have to show that $g_{2}^{- 1} g_{1} \in H$ which is equivalent to showing:

(g_{2}^{- 1} g_{1}) h \in H

Notice that $g_{0} \in g_{1} H$ so then $n \in H$ where $g_{0} = g_{1} n$ . Further, notice $g_{0} \in g_{2} H$ so then there is some $m \in H$ where $g_{0} = g_{2} m$ . Thus:

g_{1} n = g_{2} m

So notice that:

g_{2}^{- 1} g_{1} = m n^{- 1} \in H

which is what we wanted! Thus $g_{1} H \subseteq g_{2} H$ . Without loss of generality, we can do the same argument with $g_{1}$ and $g_{2}$ swapped, so then $g_{2} H \subseteq g_{1} H$ . Thus $g_{1} H = g_{2} H$ .

☐

This is a different method of proving things we saw in Left (and Right) Cosets form a Partition in G (you can see the similarities, but different lemmas are used here).