16 - The Four Horsemen Theorems

Last time we talked about thee first isomorphism 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

Let's use a different example to look at the other 3 theorems we'll cover.

Foundational Example

Consider $G = \frac{Z}{24 Z}$ . Recall that because it's Cyclic Group via Cyclic Groups of the Same Order are Isomorphic, then the subgroups of $\frac{Z}{24 Z}$ is 1:1 with the divisors of 24. Let's look at the lattice.

Notice how the lattice makes a literal diamond between subgroups. The next theorem covers when these cases happen.

Theorem

Suppose $A, B \leq G$ and $A \leq N_{G} (B)$ (see Normalizer, so every element of $A$ normalizes $B$ , so $a B a^{- 1} = B$ for all $a \in A$ ). Then:

$A B = ⟨ A \cup B ⟩$ is the smallest subgroup containing $A$ and $B$ . It is the "tip" of the diamond in the lattice.
$A \cap B ⊴ A$ and $B ⊴ A B$
$\frac{A B}{B} ≅ \frac{A}{A \cap B}$

This may be the most useless theorem ever, but the last finding is really important. Namely, for when you want to "cancel" $B$ from $A B$ then you have to consider the factor of $A \cap B$ (ie: you get some elements of $B$ that "bleed over")

Note

If $B ⊴ G$ (see Normal Subgroup) then $N_{G} (B) = G$ so $A \leq N_{G} (B)$ .

Example

Here $A = ⟨ 6 ⟩$ and $B = ⟨ 4 ⟩$ . Then $A B = ⟨ 2 ⟩$ and $A \cap B = ⟨ 12 ⟩$ .

If you take $A B$ and divide it into the cosets, you can make pairings between cosets of $A B$ versus that of just $A$ w.r.t $A \cap B$ .

Proof

Proof

(Sketch of (3)): Define a map $φ : A \to \frac{A B}{B}$ such that $a \mapsto a B$ (note that $a B \in \frac{A B}{B}$ still, our notation still makes sense). It's a Homomorphism (or Group Morphism) since $a_{1} B \cdot a_{2} B = a_{1} a_{2} B$ so $φ (a_{1} a_{2}) = φ (a_{1}) φ (a_{2})$ .

Note that $φ$ is surjective since if you take any $(a b) B$ for some $a b \in A B$ , then check that:
$(a b) B = a B$
since $a^{- 1} (a b) = b \in B$ via Left (and Right) Cosets form a Partition in G. Then $φ (a) = a B = (a b) B$ so then $φ$ is surjective.

Notice also $Ker (φ) = {a \in A | a B = 1 B}$ . But $a B = 1 B ⟺ a \in B$ , so then:
$Ker (φ) = {a \in A | a \in B} = A \cap B$
showing injectivity.

Using the The First Isomorphism Theorem of Groups, The Fundamental Theorem for Group Morphisms, then:
$\frac{A}{Ker (φ)} = \frac{A}{A \cap B} ≅ \frac{A B}{B}$
☐

Theorem

Suppose $H, K ⊴ G$ (see Normal Subgroup) and $H \leq K$ . Then:

Since $\frac{K}{H} = {k H | k \in K}$ (see Quotient Group) then using $π_{H} : G \to \frac{G}{H}$ such that $g \mapsto g H$ has $π_{H} (K) = \frac{K}{H}$ and then $\frac{K}{H} ⊴ \frac{G}{H}$ .
$\frac{(\frac{G}{H})}{\frac{K}{H}} ≅ \frac{G}{K}$

Example

Like for the other theorems consider $\frac{Z}{24 Z}$ .

Here $H = ⟨ 8 ⟩$ and $K = ⟨ 4 ⟩$ . Here:
$(\frac{\frac{(\frac{Z}{24 Z})}{H}}{\frac{K}{H}}) = {(0 + H) + \frac{K}{H}, (1 + H) + \frac{K}{H}, \dots, (3 + H) + \frac{K}{H}}$
Which equals:
$\frac{(\frac{Z}{24 Z})}{K} = the same thing$

Proof

Proof

The strategy is to find a surjection $φ : \frac{G}{H} \to \frac{G}{K}$ with $Ker (φ) = \frac{K}{H}$ defined by $g H \mapsto g K$ . If you're worried about well-definedness, it turns out this will just work.

☐

Finally for the last one:

Theorem

Suppose $K ⊴ G$ (see Normal Subgroup) with $π : G \to \frac{G}{K}$ be the cannonical projection. Then:

$A \leq B$ in $G$ iff $\overset{―}{A} \leq \overset{―}{B}$ in $\frac{G}{K} = \overset{―}{G}$
$A ⊴ G$ implies $\overset{―}{A} ⊴ \overset{―}{G}$ . This becomes an iff when we have subgroups $H \leq G$ that contain $K$ .
Then assuming $H \leq G$ that contain $K$ :
$A \leq B$ then $| B : A | = | \overset{―}{B} : \overset{―}{A} |$ (see Index (Groups))
$\overset{―}{A \cap B} = \overset{―}{A} \cap \overset{―}{B}$
$\overset{―}{⟨ A \cup B ⟩} = ⟨ \overset{―}{A} \cup \overset{―}{B} ⟩$

Example

Consider $D_{8}$ :

If we alternatively look at $K = ⟨ r^{2} ⟩ = {1, r^{2}}$ then the left cosets of $K$ in $D_{8}$ are:

Notice we don't care which representative we use for each subgroup of cosets, so then we can use the bar notation to indicate a coset:

$\overset{―}{1} = \overset{―}{r^{2}}$
$\overset{―}{r} = \overset{―}{r^{3}}$
$\overset{―}{s} = \overset{―}{s r^{2}}$
$\overset{―}{s r} = \overset{―}{s r^{3}}$
which all are elements of $\frac{D_{8}}{K} = {\overset{―}{1}, \overset{―}{r}, \overset{―}{s}, \overset{―}{s r}}$ . But look! We can make some map:
${1, s, s r^{2}, r^{2}} \to {\overset{―}{1}, \overset{―}{s}, \overset{―}{s r^{2}}, \overset{―}{r^{2}}} = {\overset{―}{1}, \overset{―}{s}}$
namely, use $π : D_{8} \to \frac{D_{8}}{K}$ such that $g \mapsto g K = \overset{―}{g}$ . Then for each normal subgroup in the lattice (see the orange boxes in the picture of the lattice):
$H_{1} = {1, s, s r^{2}, r^{2}}$ has $π (H_{1}) = \overset{―}{H_{1}}$
$H_{2} = {1, r, r^{2}, r^{3}} \to {\overset{―}{1}, \overset{―}{r}, \overset{―}{r^{2}}, \overset{―}{r^{3}}} = {\overset{―}{1}, \overset{―}{r^{2}}} = \overset{―}{H_{2}}$ .
... we can repeat with each normal subgroup, and we'll get $π (H_{i}) = \overset{―}{H_{i}}$ .
$H_{3} = {1, s r^{3}, r^{2}, s r} \to {\overset{―}{1}, \overset{―}{s r^{3}}, \overset{―}{r^{2}}, \overset{―}{s r}} = {\overset{―}{1}, \overset{―}{s r}} = \overset{―}{H_{3}}$
But some of them became the other normal subgroups. Namely $π ({1, s}) = \overset{―}{H_{1}}$
$π {1, s r^{2}} = \overset{―}{H_{4}}$
...

We can make a lattice for what becomes what, and what stays the same:

Which is the same lattice picture we had before, but only with the normal subgroups!

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!

☐

Let's use a different example to look at the other 3 theorems we'll cover.

Foundational Example

!16 - The Four Horsemen Theorems 2025-02-13 07.15.14.excalidraw

Notice how the lattice makes a literal diamond between subgroups. The next theorem covers when these cases happen.

!The Second Isomorphism Theorem (The Diamond Iso. Theorem)

!The Third Isomorphism Theorem (The Double Fraction Theorem)

Finally for the last one:

!The Fourth Isomorphism Theorem (The Lattice Isomorphism Theorem)