Coset Product

coset product

We define the coset product for $H \leq G$ as:

(g_{1} H) \cdot (g_{2} H) := (g_{1} g_{2}) H

for all $g_{1}, g_{2} \in G$ . Then this operation is well defined iff $H ⊴ G$ .

We want to prove that this is well defined; namely if $g_{1} H = \tilde{g_{1}} H$ and $g_{2} H = \tilde{g_{2}} H$ do we have:

(g_{1} g_{2}) H = (\tilde{g_{1}} \tilde{g_{2}}) H

Example

To highlight that it is not always well-defined, consider the example $H = {1, (1 3)}$ for $G = S_{3}$ . We had:

Notice $(1 2) H \cdot (2 3) H = (1 2) (2 3) H = (1 2 3) H$ . Namely if the green coset is $C_{2}$ and the red one is $C_{3}$ then does it always equal $C_{3}$ (as it did in our calculation).

What we see is that $(1 3 2) H \cdot (2 3) H = (1 3 2) (2 3) H = (1 3) H$ is in $C_{1}$ instead, which is different. Thus we needed the other condition of being a Normal Subgroup to be actually well defined.

The Proof

Let's do the proof using the importance of Normal Subgroup:

Proof

( $\Rightarrow$ ): We want to show that $H ⊴ G$ . Take any $h \in H$ and $g \in G$ . We got to show that $g h g^{- 1} \in H$ .

First notice that $g h \in g H$ . That means that since $g \in g H$ then $(g h) H = g H$ are the same cosets. But then because $g_{1} H = g_{2} H ⟺ g_{2}^{- 1} g_{1} \in H ⟺ g_{1}^{- 1} g_{2} \in H$ , then we can compare $(g h) H$ and $g H$ . Namely:

(g h) H \cdot g^{- 1} H = (g h g^{- 1}) H

and:

g H \cdot g^{- 1} H = (g g^{- 1}) H = H

Thus $(g h g^{- 1}) H = H$ so then $g h g^{- 1} \in H$ , as we wanted.

( $\Leftarrow$ ): TBD

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