14 Lagrange's Theorem

Last time we talked about:

Left and Right Cosets
Lagrange's Divisibility Theorem of Order of Subgroups
Note that if $K ⊴ G$ (see Normal Subgroup) then $\frac{G}{K}$ is a Group, with operation defined by the Coset Product.

Some examples include:

$G = Z$ . Since it's an Abelian Group then automatically we get that all subgroups are normal. Using $K = 4 Z$ gives the quotient group $\frac{Z}{4 Z} = {0 + 4 Z, 1 + 4 Z, 2 + 4 Z, 3 + 4 Z}$ with operation $a + 4 Z + b + 4 Z = (a + b) + 4 Z$ . It comes with the projection Homomorphism (or Group Morphism) $π : Z \to \frac{Z}{4 Z}$ where $n \mapsto n + 4 Z$ . This is just our usual group $\frac{Z}{4 Z}$ .
If $G = D_{8}$ with $K = Z (D_{8})$ (see Center) where oomputing gives $K = {1, r^{2}} = ⟨ r^{2} ⟩$ , then here the center is always normal. Then $\frac{G}{K}$ is a group of order:

| \frac{G}{K} | = \frac{| G |}{| K |} = \frac{8}{2} = 4

using Lagrange's Divisibility Theorem of Order of Subgroups.

We could ask: Is $\frac{D_{8}}{K}$ a Cyclic Group? We can check by cranking out the orders and seeing if they are all equal via Orders of Cyclic Subgroups, in this case if they have an element of order $4$ since our group $\frac{G}{K}$ had 4 things in it: (we discussed this in 31 Quotient Groups and Homomorphisms#5)

$| 1 K | = 1$ since it's already the identity
$| r K | = 2$ since $(r K)^{2} = (r^{2} K) = (1 K)$ .
$| s K | = 2$ since $(s K)^{2} = s^{2} K = (1 K)$ .
$| s r K | = 2$ since $(s r K)^{2} = (s r)^{2} K = s s r^{- 1} r K = 1 K$
Thus we know that this group is not cyclic.

Now we could ask: Is $\frac{D_{8}}{K}$ isomorphic to a familiar group? This is a big question we'll tackle later in the future.

Another Example

Have $G = S_{3}$ with $K = ⟨ (1 2 3) ⟩ = {1, (1 2 3), (1 3 2)}$ . We can generate the left cosets:

$1 K = {1, (1 2 3), (1 3 2)} = C_{1}$
$(1 2 3) K = (1 3 2) K = C_{1}$ . Notice applying the permutations from $K$ onto $K$ gives the same $C_{1}$ .
The last other coset is $(1 2) K = {(2 3), (1 2), (1 3)} = C_{2}$ which is equal to the other $(1 3) K = (2 3) K = C_{2}$ .

New Theorem

Theorem

Suppose $φ : G \to H$ is a Homomorphism (or Group Morphism). Then:

$Ker (φ) ⊴ G$
$im (φ) \leq H$
The map $\frac{G}{K} \to im (φ)$ where $g K \mapsto φ (g)$ is an Isomorphism, where $K = Ker (φ)$ . Thus $\frac{G}{Ker (φ)} ≅ im (φ)$

An Example

This will be the most common way to understand a Quotient Group. To highlight this, consider the example $G = D_{8}$ . We claim that $\frac{D_{8}}{K}$ where in this case $K = ⟨ r^{2} ⟩$ then has:
$\frac{D_{8}}{K} ≅ Z_{2} \times Z_{2}$
The idea is to find some surjective morphism $φ : D_{8} \to Z_{2} \times Z_{2}$ in which $Ker (φ) = K$ .

Know that $D_{8}$ is generated by $r, s$ , so $φ$ is determined by $φ (r)$ and $φ (s)$ . We also know the key relations are that $r^{4} = 1 = s^{2}, r s = s r^{3}$ . Thus if we want such a morphism $φ$ then:
$φ (r^{4}) = 4 φ (r) = 0 = (\overset{―}{0}, \overset{―}{0})$

When we look at the codomain of

φ

its in terms of adding $\mathbb

s, hence multiplying by 4.

Further using the other relations:
$φ (s^{2}) = 2 φ (s) = 0, φ (r) + φ (s) = φ (s) - φ (r)$
Lastly, we want $φ$ to be surjective and $Ker (φ) = {1, r^{2}}$ .
$Z_{2} \times Z_{2} = {(0, 0), (0, 1), (1, 0), (1, 1)}$
The idea here is to treat the numbers as binary numbers. Namely $r \mapsto (0, 1)$ and $s \mapsto (1, 0)$ as they are flags for if we have an extra $r, s$ from the Left and Right Cosets of $K$ .

Notice here $φ$ is surjective as we wanted. We can make a table to check:

1	$r$	$r^{2}$	$r^{3}$	$s$	$s r$	$s r^{2}$	$s r^{3}$
(0,0)	(1,0)	(0,0)	(1,0)	(0,1)	(1,1)	(0,1)	(1,1)
Notice:

The $1 K$ coset had $φ (1 K) = (0, 0)$ .
The other outcomes got partitioned by their cosets.
Thus we got $Ker (φ) = K$ as desired.

But notice that in this case we really could've used a whole bunch of other maps since we only have $mod 2$ (some arithmetic just works out).

Proof

Proof

The first two will be quick to check, so let's look at (3). Define $ψ : \frac{G}{Ker (φ)} \to im (φ)$ by $ψ (g Ker (φ)) = φ (g)$ .

Let's see if $ψ$ is well defined. Suppose $g_{1} Ker (φ) = g_{2} Ker (φ)$ . Then using Left (and Right) Cosets form a Subgroup, then $g_{2}^{- 1} g_{1} \in Ker (φ)$ . Thus $φ (g_{2}^{- 1} g_{1}) = 1$ . Then $φ (g_{2})^{- 1} φ (g_{1}) = 1_{H}$ being a morphism. Thus $φ (g_{1}) = φ (g_{2})$ showing it's well defined.

We should check $ψ$ is a morphism:
$\begin{aligned} ψ ((g_{1} Ker (φ)) (g_{2} Ker (φ))) & = ψ (g_{1} g_{2} Ker (φ)) \\ = φ (g_{1} g_{2}) \\ = φ (g_{1}) φ (g_{2}) \\ = ψ (g_{1} Ker (φ)) ψ (g_{2} Ker (φ)) \end{aligned}$
Now is $ψ$ surjective? Let $h \in im (φ)$ , then there is some $g \in G$ where $φ (g) = h$ . But then $ψ (g Ker (φ)) = φ (g) = h$ .

Now is $ψ$ injective? Notice:
$\begin{aligned} Ker (ψ) & = {g Ker (φ) \in \frac{G}{Ker (φ)} | φ (g) = 1_{H}} \\ = {g Ker (φ) | g \in Ker (φ)} \\ = {g Ker (φ) | g Ker (φ) = 1 Ker (φ)} \\ = {1 Ker (φ)} \\ = {1_{\frac{G}{Ker (φ)}}} \end{aligned}$
so $ψ$ is injective. Thus $ψ$ is an isomorphism between the two groups!

☐