MIT 18.701 — Lecture 14

Icosahedron Symmetries, Jordan-Holder

The only group we've shown to be simple is $C_p$ for primes $p$ . Let's find a more complicated simple group.

§ Icosahedron Symmetries are Simple

Consider $G$ to be the rotational symmetry group of a regular icosahedron.

We know from Lecture 12 that $G \subseteq SO_3(\mathbb{R})$ has $60$ group elements. These elements are:

\text{group elements of } G = \begin{cases} \text{identity rotations} & (+1) \\ \text{rotations around an axis through a face} & (+20) \\ \text{rotations around an axis through an edge} & (+15) \\ \text{rotations around an axis through a vertex} & (+24) \end{cases}

What do the conjugacy classes of $G$ look like? Well, conjugation is just a change of coordinates, so:

\text{conjugacy classes of } G = \begin{cases} \text{identity rotations} & (+1) \\ \text{rotations around an axis through a face} & (+20) \\ \text{rotations around an axis through an edge} & (+15) \\ \text{rotations by $\pm \frac{2\pi}{5}$ about an axis through a vertex} & (+12) \\ \text{rotations by $\pm \frac{4\pi}{5}$ about an axis through a vertex} & (+12) \\ \end{cases}

The above will let us show $G$ is simple!

Theorem. (Simplicity of Icosahedra) The group $G$ is simple.

Proof: Suppose $H \unlhd G$ is a normal subgroup. Recall this means $gHg^{-1} = H$ for all $g \in G$ . Equivalently, $H$ must be the union of conjugacy classes of $G$ .

But by inspection, the union of conjugacy classes of $G$ (including $\{1\}$ ) cannot have a number of elements that is a divisor of $60$ (other than $1$ or $60$ ). So $H$ cannot exist. $\blacksquare$

The crux of the above argument, to reiterate, was the following useful observation:

Theorem. (Normals are Conjugacy Classes) If $H \unlhd G$ , then $H$ is the union of conjugacy classes of $G$ , including the conjugacy class consisting of just the identity.

Remark. It turns out that $G \cong A_5$ . One way to see this is to note:

The rotations $G$ of a regular icosahedron are isomorphic to the rotations $G'$ of a regular dodecahedron.
There are $5$ cubes inscribed in a regular dodecahedron.

Thus, $G$ is isomorphic to permutations of those $5$ cubes, and it turns out that these permutations must be even.

Note that point #2 also gives a good strategy for drawing a dodecahedron…

§ Composition Series

Definition (Composition Series). Let $G$ be a group. Then a composition series of $G$ is a chain:

G = G_0 \supseteq G_1 \supseteq G_2 \supseteq \dots \supseteq G_r = \{1\}

satisfying the property that for all $i$ , we have $G_i \unlhd G_{i - 1}$ and $G_{i - 1} / G_i$ is simple.

Remark. Confusingly, smaller groups get larger indices. Also, simple groups are not trivial, so $G_{i - 1} \neq G_i$ .

Example. The group $C_6 = \langle x \rangle$ has two composition series:

\langle x \rangle \supseteq \langle x^3 \rangle \supseteq \{1\} ~~ \text{ and } ~~ \langle x \rangle \supseteq \langle x^2 \rangle \supseteq \{1\}

Meanwhile, the infinite group $(\mathbb{Z}, +)$ has no composition series: nontrivial subgroups of $\mathbb{Z}$ are isomorphic to $\mathbb{Z}$ .

Theorem. (Existence of Composition Series) Every finite group $G$ has a composition series.

Proof: It's good enough to build the following chisel.

Chisel. Given groups $G$ and $H$ with $H \unlhd G$ and $G/H$ not simple,

the chisel constructs a group $T$ such that $H \unlhd T \unlhd G$ , with $H \neq T \neq G$ .

To build a composition series given a chisel, we can just start with $\{1\} \unlhd G$ and use the chisel repeatedly.

How does the chisel work? Well, since $G / H$ is not simple, there must exist some normal subgroup $N \unlhd G / H$ besides $\{1\}$ and $G / H$ . So the chisel decides to choose $T := \pi^{-1}(N)$ , where $\pi: G \to G / H$ is the quotient map.

We now check the chisel's work by showing this choice of $T$ is valid:

We know $H$ is a subgroup of $T$ and $T$ is a subgroup of $G$ both by the Correspondence Theorem.
The fact $H \unlhd T$ follows from directly combining $H \unlhd G$ and $T \subseteq G$ together. Indeed,
$\text{``For all } g \in G, \text{ we have } gHg^{-1} = H\text{.''} ~~~~~ \text{ implies } ~~~~~ \text{``For all } t \in T, \text{ we have } tHt^{-1} = H\text{.''}$
The fact $T \unlhd G$ follows from the property $\pi(T) = N \unlhd G / H$ . Indeed, for any $g \in G$ ,
$\pi(gTg^{-1}) = \pi(g) \pi(T) \pi(g^{-1}) = \pi(g) N \pi(g)^{-1} = N \text{ since } N \unlhd G / H.$
Thus, $gTg^{-1} \subseteq \pi^{-1}(N) = T$ for all $g$ . And $gTg^{-1} \subseteq T$ for all $g \in G$ is good enough to show $T \unlhd G$ .

So the chisel works correctly, and we're done. $\blacksquare$

§ The Jordan-Holder Theorem

It turns out composition series are unique, up to order of “prime factors”!

Theorem. (Jordan-Holder) Suppose $G$ has two composition series:

G = A_0 \supseteq A_1 \supseteq \dots \supseteq A_r = \{1\} ~~ \text{ and } ~~ G = B_0 \supseteq B_1 \supseteq \dots \supseteq B_s = \{1\}

Then the simple quotients $\{A_i / A_{i + 1}: i = 0, 1, \dots, r - 1\}$ map bijectively with $\{B_i / B_{i + 1} : i = 0, 1, \dots, s - 1\}$ .

Example. The two composition series of $C_6$ have the same sets of simple quotients: $\{C_3, C_2\}$ and $\{C_2, C_3\}$ .

\begin{align*}& \langle x \rangle \supseteq \langle x^3 \rangle \supseteq \{1\} ~~ \text{ has quotients } ~~ \langle x \rangle / \langle x^3 \rangle \cong C_3 \text{ and } \langle x^3 \rangle / \{1 \} \cong C_2. \\ & \langle x \rangle \supseteq \langle x^2 \rangle \supseteq \{1 \} ~~ \text{ has quotients } ~~ \langle x \rangle / \langle x^2 \rangle \cong C_2 \text{ and } \langle x^2 \rangle / \{1 \} \cong C_3.\end{align*}

We'll spend the rest of lecture proving the Jordan-Holder Theorem. But first, some precautions:

Two non-isomorphic groups can have the same sets of simple quotients. For example,
$\begin{align*}& S_3 \supseteq A_3 \supseteq \{1\} ~~ \text{ has quotients } ~~ S_3 / A_3 \cong C_2 \text{ and } A_3 / \{1\} \cong C_3. \\ & C_6 \supseteq C_3 \supseteq \{1\} ~~ \text{ has quotients } ~~ C_6 / C_3 \cong C_2 \text{ and } C_3 / \{1\} \cong C_3.\end{align*}$
But unfortunately, $C_6 \not \cong S_3$ .
The condition $G_i \unlhd G_{i - 1}$ does not imply, transitively, that $G_m \unlhd G_n$ for all $m \geq n$ . For example, taking
$G_1 = S_4 \hspace{0.8cm} G_2 = \{1, ( 1\ 2)(3 \ 4), (1 \ 3)(2 \ 4), (1 \ 4)(2 \ 3)\} \hspace{0.8cm} G_3 = \{1, (1 \ 2)(3 \ 4)\},$
we see that $G_2 \unlhd G_1$ and $G_3 \unlhd G_2$ both hold, but $G_3 \unlhd G_1$ does not hold.

§ The Second Isomorphism Theorem

We'll need two lemmas to prove the Jordan-Holder Theorem. This one's the first.

Theorem. (The Second Isomorphism Theorem) Suppose $H, K \subseteq G$ are subgroups with $K \unlhd G$ . Then:

The set $HK = \{hk \mid h \in H, k \in K\}$ is a subgroup of $G$ .
The isomorphism $HK / K \cong H / (H \cap K)$ holds. (Here, we are also implicitly claiming that $(H \cap K) \unlhd H$ .)

Proof: To prove the first point, we simply check that $1 \in HK$ (obviously), that $(hk)^{-1} = h^{-1}(hk^{-1}h^{-1}) \in HK$ , and:

hk h' k' = \underbrace{(hh')}_{\in H} \ \underbrace{(h'^{-1}k h')}_{\in K} \ \underbrace{(k')}_{\in K} \in HK.

For the second point, we'll use the quotient map $\pi: G \to G / K$ . Then:

The First Isomorphism Theorem says $HK / K \cong \pi(HK)$ .
Since $K = \ker \pi$ , we have $\pi(HK) = \pi(H)$ .
Well, $\pi(H)$ is itself the image of $\pi_{|H}: H \to G / K$ . But $\ker(\pi_{|H}) = H \cap K$ , so $\pi(H) \cong H / (H \cap K)$ .

Stringing together all three parts of the above yields $HK / K \cong H / (H \cap K)$ , as desired. $\blacksquare$

Some comments on the Second Isomorphism Theorem:

(#1)
Sometimes the Second Isomorphism Theorem is called the Third Isomorphism Theorem.
(#2)
For an alternative proof of the first point, rephrase the first point more naturally like so:

If $K \unlhd G$ , then the group generated by $H$ and $K$ is $HK = \{hk \mid h \in H, k \in K\}$ .

Usually, the group generated by $H$ and $K$ looks like $\{h_1k_1h_2k_2\dots \mid h_i \in H, k_i \in K\}$ . However, when $K \unlhd G$ , we have the property that for any $h \in H$ and $k \in K$ , there is some $k' \in K$ such that $kh = hk'$ .

So we may repeatedly use this property to shift all the $k_i$ 's in the product $h_1k_1h_2k_2\dots$ to the right.
(#3)
Nothing in $G$ matters besides elements in $HK$ . Thus, instead of mandating $K \unlhd G$ , we could have gotten away with only mandating that $hKh^{-1} = K$ for all $h \in H$ .

In other words, the clause “ $K \unlhd G$ ” in the theorem statement could have been replaced with “ $H \subseteq N(K)$ ”. Here, $N(K)$ denotes the normalizer of $K$ : the set $N(K) := \{g \in G: gKg^{-1} = K\}$ .

§ The Diamond Lemma

Now for the second of the two lemmas.

Theorem. (Diamond Lemma) If $H$ and $K$ are distinct normal subgroups of $G$ with $G / H$ and $G / K$ both simple, then

G / H \cong K / (H \cap K) ~ \text{ and } ~ G / K \cong H / (H \cap K).

Proof: By symmetry, it suffices only to prove $G / K \cong H / (H \cap K)$ . Well, recall that the Second Isomorphism Theorem says $H / (H \cap K) \cong HK / K$ . So it suffices to prove $G / K \cong HK / K$ .

Since $H$ and $K$ are normal subgroups of $G$ , so is $HK$ . Then $HK \unlhd G$ implies $(HK / K) \unlhd (G / K)$ as well. But $G / K$ was assumed to be simple! Thus, either $HK / K \cong \{1\}$ or $HK / K \cong G / K$ .

The latter is what we want, so assume the former: $HK / K \cong \{1\}$ , or equivalently, $H \subseteq K$ . This lets us talk about $K / H$ ; in particular, $(K / H) \unlhd (G / H)$ holds. But we now have another problem: $G / H$ was assumed to be simple!

Thus, either $K / H = \{1\}$ or $K / H = G / H$ .

The former implies $K = H$ , contradicting the assumption $K \neq H$ in the theorem statement.
The latter implies $K = G$ , contradicting the assumption that $G / K$ was simple. (Trivial groups aren't simple!)

So we're done. $\blacksquare$

There's not much to say about the Diamond Lemma. The following image reveals the meaning behind its name:

§ Proof of The Jordan-Holder Theorem

We're now ready to prove the Jordan-Holder Theorem. Recall that we have a group $G$ with two composition series:

G = A_0 \supseteq A_1 \supseteq \dots \supseteq A_r = \{1\} ~~ \text{ and } ~~ G = B_0 \supseteq B_1 \supseteq \dots \supseteq B_s = \{1\}

We'd like to show these two series are equivalent; that is, their simple quotients $\{A_i / A_{i + 1}\}$ and $\{B_i / B_{i + 1}\}$ map bijectively with each other. Our approach is induction on $r$ , with inductive hypothesis:

Inductive Hypothesis. If a group $G'$ has any composition series of length $r - 1$ ,
then all composition series of $G'$ are equivalent.

(Here we WLOG $r \geq s$ .) The entire rest of the proof is captured in the following chain of equivalences:

There are two steps to the proof: step 1 and step 2.

1
We may assume $A_1 \neq B_1$ ; indeed, if $A_1 = B_1$ , then the inductive hypothesis would tell us $A_1 \supseteq \dots \supseteq A_r$ and $B_1 \supseteq \dots \supseteq B_s$ were equivalent, and we'd be done.

Now consider $K_1 = A_1 \cap B_1$ , along with any composition series $K_1 \supseteq \dots \supseteq K_t$ of $K_1$ .

Claim. The chain $A_1 \supseteq K_1 \supseteq \dots \supseteq K_t$ is a valid composition series of $A_1$ .

Proof: It suffices to show $A_1 / K_1$ is simple. This follows from the Diamond Lemma, which says $A_1 / K_1 \cong G / B_1$ . And $G / B_1$ is simple by definition. $\square$

Thus, we have constructed two composition series of $A_1$ (green in the original diagram):
$A_1 \supseteq A_2 \supseteq \dots \supseteq A_r ~~ \text{ and } ~~ A_1 \supseteq K_1 \supseteq \dots \supseteq K_t.$
But the former of these two composition series has length $r - 1$ . So by the inductive hypothesis, these two composition series are equivalent, as desired.
2
This follows immediately from the Diamond Lemma, which says $G / A_1 \cong B_1 / K_1$ , and $G / B_1 \cong A_1 / K_1$ .

And that's the proof! $\blacksquare$