Inverse element of a quotient group (proof)

Question

When we prove that the quotient group is indeed a group we have for the inverse element, if $G$ is a group and $H$ is a normal subgroup of $G$

$$(xH)(x^{-1}H)=xx^{-1}HH=xx^{-1}H=1H$$

We were told that the fact in the first equality that we can switch the places is because $H$ is a normal subgroup. I don't see the connection unfortunately.

For the normal subgroup we have that $xHx^{-1} \subseteq H$ but how does that relate to commutativity, if it is not given that $G$ is abelian?

Answer 1

Firstly, the internal product of subsets and elements of a group is associative, so we can unambiguously write the product as $\left(x H\right) \left(x^{-1} H\right) = x H x^{-1} H$ without using parentheses; or, we could use any other valid set of parentheses—like in $x \left(H x^{-1}\right) H$—if we want.

If $H$ is a normal subgroup, what can we say about the right-coset, $H x^{-1}$, in the middle of the product? We can substitute something in its place that can help you understand why the equality in the question holds.

Answer 2

If $H$ is a subgroup of $G$, then the cosets $\{xH \mid x \in G\}$ are subsets of $G$. But if $H$ is additionally a normal subgroup, the cosets can be treated like objects with their own multiplication. The normality condition $x H x^{-1} \subseteq H$ is exactly what makes this well-defined.

The set $(xH)(x^{-1} H)$ is the set of all $(xh_1)(x^{-1}h_2)$, where $h_1$ and $h_2$ range over $H$. Since $H$ is normal, we know that $x h_1 x^{-1} \in H$. Then by associativity, and the fact that $H$ is a closed under multiplication: $$ (xh_1)(x^{-1}h_2) = (xh_1x^{-1}) h_2 \in H $$ Therefore $(xH)(x^{-1} H) \subseteq H$. On the other hand, given $h \in H$, $$ h = e h = x x^{-1} h = xe x^{-1} h = (xe)(x^{-1} h) \in (xH)(x^{-1}H) $$ Therefore $H \subseteq (xH)(x^{-1} H)$.

Answer 3

At some point you will have seen the following

Lemma: A subgroup $N$ of a group $G$ is normal iff for all $g\in G$, we have $$gN=Ng.$$

Equivalent definitions of normal subgroups

Since $G$ is associative, we have for all $x\in G$ and $H$ normal in $G$, $$(xH)(x^{-1}H)=x(Hx^{-1})H,$$ but now the Lemma gives $Hx^{-1}=x^{-1}H$. Can you continue from here?