12 Quotient Groups

left/right cosets

For any $N \leq G$ (see Subgroup) and any $g \in G$ let:
$g N = {g n | n \in N}; N g = {n g | n \in N}$
be called (respectively) a left coset and a right coset of $N$ in $G$ . Any element of a coset is called a representative for the coset.

We denote the set of left cosets by $\frac{G}{H}$ , as we see in Quotient Group.

Warning

In general $g H \neq H g$ for all $g \in G$ . Essentially, once the left/right cosets agree with each other, then we have a Normal Subgroup specifically for that $H$ .

Examples

Say we have $G = S_{3}$ (see Symmetric Group & Permutations) and $H = ⟨ (1 2) ⟩ {1, (1 2)}$ . Then the left-cosets of $H$ are;

$g \in G$	$g H$
1	${1 1, 1 (1 2)} = {1, (1 2)} = H$
$(1 2)$	${(1 2) 1, (1 2) (1 2)} = {(1 2), 1}$
$(1 3)$	${(1 3), (1 2 3)}$
$(2 3)$	${(2 3), (1 3 2)}$
$(1 2 3)$	${(1 2 3), (1 3)}$
$(1 3 2)$	${(1 3 2), (2 3)}$
Notice that these are partitioned. Namely name:

$1 H, (1 2) H = C_{1}$
$(1 3) H, (1 2 3) H = C_{2}$
$(2 3) H, (1 3 2) H = C_{3}$
creating a partition of 3 elements.

Note

For one right coset see the order matters:
$H (1 3) = {1 (1 3), (1 2) (1 3)} = {(1 3), (1 3 2)} \neq (1 3) H$

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).

Lemma

For any $H \leq G$ (see Subgroup) and $g \in G$ we have:
$g H = H g ⟺ g H g^{- 1} = H$
(see Left and Right Cosets) In fact, the following are equivalent:

$g H = H g$ for any $g \in G$
$g H g^{- 1} = H$ for all $g \in G$
$g H g^{- 1} \subseteq H$ for all $g \in G$
$g h g^{- 1} \in H$ for all $g \in G$
$N_{G} (H) = G$ (see Normalizer)
$H ⊴ G$ (see Normal Subgroup) (comes directly from (5))

Proof

For the former, we'll do both directions:

$(\Rightarrow)$ : Suppose $g H = H g$ . Take any $h \in H$ where then:
$g h \in g H = H g \Rightarrow g h \in H g$
so then there is some $h^{'} \in H$ where $g h = h^{'} g$ . Thus:
$g h g^{- 1} = h^{'} \in H$
thus proving (3), namely $g H g^{- 1} \subseteq H$ .

We just need to prove that $H \subseteq g H g^{- 1}$ . So take any $h \in H$ . We want to show (want to find) some $h^{'} \in H$ such that $h = g h^{'} g^{- 1}$ . So then we'd want $h = g h^{'} g^{- 1} \Leftarrow g^{- 1} h g$ .

So choose $h^{'} := g^{- 1} h g$ . Now notice that $h g \in H g = g H$ . Thus then $h g = g \tilde{h}$ for some $\tilde{h} \in H$ . Then:
$h^{'} = g^{- 1} h g = g^{- 1} (g \tilde{h}) = \tilde{h} \in H$
as we needed to verify. Thus then:
$\Rightarrow h = g (g^{- 1} h g) g^{- 1} \in g H g^{- 1} \Rightarrow H \subseteq g H g^{- 1}$

☐

normal

Let $H \leq G$ be normal if:
$N_{G} (H) = G$
(see Normalizer). Which means you get the equivalence of Equivalences for Normal Subgroups.

We'll use the notation $K ⊴ G$ when $K$ is normal to $G$ .

Example

Recall our Left and Right Cosets#Examples $G = S_{3}$ with $H = ⟨ (1 2) ⟩ = {1, (1 2)}$ . Then we saw that:
$(1 3) H \neq H (1 3)$
thus $H$ is not normal. On the other hand if we have:
$K = ⟨ (1 2 3) ⟩ = {1, (1 2 3), (1 3 2)}$
is normal.