Chapter 15 - Factor-Group Computations and Simple Groups

This chapter has two interlocking themes. The first is computational: how does one actually identify a factor group $G/N$ by finding the right surjective homomorphism? The second is structural: which groups admit no nontrivial normal collapse, and why does that matter? The answer leads to simple groups, composition series, and the Jordan-Holder theorem — the group-theoretic analogue of unique prime factorization.

§15.1 The Fundamental Homomorphism Theorem as a Computational Tool

The Fundamental Homomorphism Theorem (FHT, also called the First Isomorphism Theorem) was stated in Chapter 14. Here we use it as a strategy for computing factor groups.

Theorem 15.1 (Fundamental Homomorphism Theorem). Let $\phi: G \to H$ be a surjective homomorphism with $\ker(\phi) = N$. Then

More precisely, the map $\mu: G/N \to H$ defined by $\mu(gN) = \phi(g)$ is a well-defined isomorphism.

Proof

This was proved in Chapter 14. We recall the key steps for reference.

Well-defined: If $gN = g'N$, then $g' = gn$ for some $n \in N = \ker(\phi)$, so $\phi(g') = \phi(gn) = \phi(g)\phi(n) = \phi(g)e = \phi(g)$.

Homomorphism: $\mu(gN \cdot g'N) = \mu(gg'N) = \phi(gg') = \phi(g)\phi(g') = \mu(gN)\mu(g'N)$.

Injective: If $\mu(gN) = e_H$, then $\phi(g) = e_H$, so $g \in \ker(\phi) = N$, hence $gN = N$.

Surjective: Since $\phi$ is surjective, for every $h \in H$ there exists $g \in G$ with $\phi(g) = h$, so $\mu(gN) = h$.

The strategy. To identify $G/N$:

1. Find a group $H$ that you suspect is the answer.
2. Construct a surjective homomorphism $\phi: G \to H$.
3. Verify that $\ker(\phi) = N$.
4. Conclude $G/N \cong H$ by the FHT.

This is almost always faster than listing all cosets and building the multiplication table from scratch.

§15.2 Worked Factor-Group Computations

Example 15.2: $(\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(\bar{2}, \bar{3})\rangle$

Step 1: Compute the subgroup being collapsed.

Let $N = \langle(\bar{2}, \bar{3})\rangle$ inside $G = \mathbb{Z}_4 \times \mathbb{Z}_6$. We compute:

So $|(\bar{2}, \bar{3})| = 2$ and $N = \{(\bar{0}, \bar{0}),\, (\bar{2}, \bar{3})\}$, hence $|N| = 2$.

Step 2: Determine the order of the quotient.

Step 3: Find the right homomorphism (or classify by structure).

Since $G = \mathbb{Z}_4 \times \mathbb{Z}_6$ is abelian, so is $G/N$. By the Fundamental Theorem of Finitely Generated Abelian Groups, an abelian group of order $12$ is isomorphic to either $\mathbb{Z}_{12}$ or $\mathbb{Z}_2 \times \mathbb{Z}_6$.

To distinguish them, find an element of order $12$ in $G/N$. Consider the coset $(\bar{1}, \bar{0}) + N$. We need the smallest $m > 0$ such that $m(\bar{1}, \bar{0}) \in N$. Now $m(\bar{1}, \bar{0}) = (\bar{m}, \bar{0})$. For this to be in $N = \{(\bar{0}, \bar{0}), (\bar{2}, \bar{3})\}$:

• $(\bar{m}, \bar{0}) = (\bar{0}, \bar{0})$ requires $4 \mid m$, so $m = 4$ at minimum.
• $(\bar{m}, \bar{0}) = (\bar{2}, \bar{3})$ requires $\bar{0} = \bar{3}$ in $\mathbb{Z}_6$, which is impossible.

So $(\bar{1}, \bar{0}) + N$ has order $4$ in $G/N$.

Now consider $(\bar{0}, \bar{1}) + N$. We need $m(\bar{0}, \bar{1}) = (\bar{0}, \bar{m}) \in N$:

• $(\bar{0}, \bar{m}) = (\bar{0}, \bar{0})$ requires $6 \mid m$, so $m = 6$ at minimum.
• $(\bar{0}, \bar{m}) = (\bar{2}, \bar{3})$ requires $\bar{0} = \bar{2}$ in $\mathbb{Z}_4$, which is impossible.

So $(\bar{0}, \bar{1}) + N$ has order $6$.

Since $\gcd(4, 6) = 2$, the element $(\bar{1}, \bar{0}) + N + (\bar{0}, \bar{1}) + N = (\bar{1}, \bar{1}) + N$ has order $\operatorname{lcm}(4, 6) = 12$. (One checks: $\operatorname{ord}((\bar{1}, \bar{1}) + N) = 12$ since $m(\bar{1}, \bar{1}) = (\bar{m}, \bar{m}) \in N$ forces both $4 \mid m$ and $6 \mid m$, or $\bar{m} = \bar{2}$ in $\mathbb{Z}_4$ and $\bar{m} = \bar{3}$ in $\mathbb{Z}_6$ simultaneously. The second case requires $m \equiv 2 \pmod{4}$ and $m \equiv 3 \pmod{6}$; by CRT this gives $m \equiv 9 \pmod{12}$, but then $m = 9$ gives $(\bar{1}, \bar{3}) \neq (\bar{2}, \bar{3})$. Actually, $9 \bmod 4 = 1 \neq 2$. So only the first case applies: $\operatorname{lcm}(4,6) = 12$.)

Therefore $G/N$ has an element of order $12$, so

Alternative via FHT: Define $\phi: \mathbb{Z}_4 \times \mathbb{Z}_6 \to \mathbb{Z}_{12}$ by $\phi(\bar{a}, \bar{b}) = \overline{3a + 2b}$ in $\mathbb{Z}_{12}$. One checks this is a well-defined surjective homomorphism (since $\phi(\bar{1}, \bar{1}) = \bar{5}$ has order $12$ in $\mathbb{Z}_{12}$, so the image is all of $\mathbb{Z}_{12}$). The kernel consists of $(\bar{a}, \bar{b})$ with $12 \mid 3a + 2b$ (lifting to integers). One verifies $\ker(\phi) = \{(\bar{0}, \bar{0}), (\bar{2}, \bar{3})\} = N$.

Example 15.3: $(\mathbb{Z}_2 \times \mathbb{Z}_4)/\langle(\bar{1}, \bar{2})\rangle$

Let $N = \langle(\bar{1}, \bar{2})\rangle$ in $G = \mathbb{Z}_2 \times \mathbb{Z}_4$.

Compute: $(\bar{1}, \bar{2})$, then $2(\bar{1}, \bar{2}) = (\bar{0}, \bar{0})$. So $|N| = 2$ and $N = \{(\bar{0}, \bar{0}), (\bar{1}, \bar{2})\}$.

An abelian group of order $4$ is either $\mathbb{Z}_4$ or $\mathbb{Z}_2 \times \mathbb{Z}_2$ (the Klein four-group $V_4$).

Check whether $G/N$ has an element of order $4$. The coset $(\bar{0}, \bar{1}) + N$ has order $m$ where $m$ is the smallest positive integer with $m(\bar{0}, \bar{1}) = (\bar{0}, \bar{m}) \in N$:

• $(\bar{0}, \bar{m}) = (\bar{0}, \bar{0})$ requires $4 \mid m$.
• $(\bar{0}, \bar{m}) = (\bar{1}, \bar{2})$ requires $\bar{0} = \bar{1}$ in $\mathbb{Z}_2$, impossible.

So $\operatorname{ord}((\bar{0}, \bar{1}) + N) = 4$, hence

Example 15.4: $(\mathbb{Z} \times \mathbb{Z})/\langle(1, 1)\rangle \cong \mathbb{Z}$

Let $N = \langle(1,1)\rangle = \{(n, n) : n \in \mathbb{Z}\}$ in $G = \mathbb{Z} \times \mathbb{Z}$.

Using FHT: Define $\phi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ by $\phi(a, b) = a - b$.

• Homomorphism: $\phi((a,b) + (c,d)) = \phi(a+c, b+d) = (a+c) - (b+d) = (a-b) + (c-d) = \phi(a,b) + \phi(c,d)$.
• Surjective: For any $n \in \mathbb{Z}$, $\phi(n, 0) = n$.
• Kernel: $\phi(a, b) = 0 \iff a = b \iff (a, b) = (a, a) \in \langle(1,1)\rangle$.

So $\ker(\phi) = N$, and by the FHT:

This is a beautiful example because the quotient of an uncountable-rank-looking object turns out to be the simplest infinite cyclic group.

Example 15.5: $(\mathbb{Z} \times \mathbb{Z})/\langle(2, 2)\rangle$

Define $\phi: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z} \times \mathbb{Z}_2$ by $\phi(a, b) = (a - b,\, \bar{b})$ where $\bar{b}$ is $b \bmod 2$.

• Homomorphism: routine check.
• Surjective: $\phi(n, 0) = (n, \bar{0})$ and $\phi(n, 1) = (n-1, \bar{1})$, so the image is all of $\mathbb{Z} \times \mathbb{Z}_2$.
• Kernel: $\phi(a,b) = (0, \bar{0})$ requires $a = b$ and $2 \mid b$, so $(a,b) = (2k, 2k)$ for some $k$. Thus $\ker(\phi) = \langle(2,2)\rangle$.

Therefore

Example 15.6: $\mathbb{R}^{*}/\{1, -1\} \cong \mathbb{R}_{>0}$

Let $G = \mathbb{R}^{*} = (\mathbb{R} \setminus \{0\}, \cdot)$ and $N = \{1, -1\}$.

Define $\phi: \mathbb{R}^{*} \to \mathbb{R}_{>0}$ by $\phi(x) = |x|$.

• Homomorphism: $\phi(xy) = |xy| = |x||y| = \phi(x)\phi(y)$.
• Surjective: every positive real is already in $\mathbb{R}^{*}$.
• Kernel: $\phi(x) = 1 \iff |x| = 1 \iff x = \pm 1$.

So $\ker(\phi) = \{1, -1\} = N$ and by the FHT:

§15.3 The $G/Z(G)$ Theorem

Definition 15.7. The center of a group $G$ is

$Z(G) \trianglelefteq G$

Proof. Let $z \in Z(G)$ and $g \in G$. Then $gzg^{-1} = zgg^{-1} = z \in Z(G)$, where we used $gz = zg$. So $Z(G)$ is stable under conjugation and hence normal.

Theorem 15.8 ($G/Z(G)$ Theorem). If $G/Z(G)$ is cyclic, then $G$ is abelian.

Proof

Suppose $G/Z(G)$ is cyclic, say $G/Z(G) = \langle gZ(G) \rangle$ for some $g \in G$.

Then every element of $G$ can be written as $g^k z$ for some integer $k$ and some $z \in Z(G)$.

Let $a, b \in G$ be arbitrary. Write $a = g^m z_1$ and $b = g^n z_2$ with $z_1, z_2 \in Z(G)$. Then:

$$ab = g^m z_1 \cdot g^n z_2 = g^m g^n z_1 z_2 = g^{m+n} z_1 z_2$$

where we used the fact that $z_1$ commutes with everything (including $g^n$). Similarly:

$$ba = g^n z_2 \cdot g^m z_1 = g^{n+m} z_2 z_1 = g^{m+n} z_1 z_2$$

using commutativity of $z_1, z_2$ with everything and with each other.

Therefore $ab = ba$ for all $a, b \in G$, so $G$ is abelian.

Note the logical content: if $G/Z(G)$ is cyclic, then $G$ is abelian, which means $Z(G) = G$, which means $G/Z(G)$ is trivial. So the hypothesis ”$G/Z(G)$ is cyclic and nontrivial” is impossible. The theorem is really a proof by contradiction in disguise.

Corollary 15.9. If $|G| = p^2$ for a prime $p$, then $G$ is abelian.

Proof

By the class equation and properties of $p$-groups, $|Z(G)| > 1$. Since $|G| = p^2$, we have $|Z(G)| \in \{p, p^2\}$.

If $|Z(G)| = p^2$, then $Z(G) = G$ and $G$ is abelian.

If $|Z(G)| = p$, then $|G/Z(G)| = p^2/p = p$, which is prime. A group of prime order is cyclic. So $G/Z(G)$ is cyclic, and by Theorem 15.8, $G$ is abelian. But then $Z(G) = G$, contradicting $|Z(G)| = p$. So this case is impossible.

In either case, $G$ is abelian.

This corollary shows that groups of order $p^2$ are isomorphic to either $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p \times \mathbb{Z}_p$. There are no nonabelian groups of order $p^2$.

§15.4 Simple Groups

Definition 15.10. A group $G$ is simple if $G \neq \{e\}$ and the only normal subgroups of $G$ are $\{e\}$ and $G$ itself.

The abelian case

Theorem 15.11. A finite abelian group is simple if and only if it has prime order (i.e., $G \cong \mathbb{Z}_p$).

Proof

($\Leftarrow$) If $|G| = p$ is prime, then by Lagrange’s theorem, the only subgroups are $\{e\}$ and $G$. Since $G$ is abelian, every subgroup is normal. So $G$ is simple.

($\Rightarrow$) Let $G$ be a finite abelian simple group. Pick $a \neq e$ in $G$. Since $G$ is abelian, $\langle a \rangle \trianglelefteq G$. By simplicity, $\langle a \rangle = G$, so $G$ is cyclic. If $|G| = n$ is composite, say $n = km$ with $1 < k, m < n$, then $\langle a^k \rangle$ is a proper nontrivial subgroup (it has order $m$), contradicting simplicity. So $|G|$ is prime.

Examples

• $\mathbb{Z}_p$ is simple for every prime $p$.
• $\mathbb{Z}_6$ is not simple: $\{0, 2, 4\} \cong \mathbb{Z}_3$ is a proper nontrivial normal subgroup.
• $A_n$ is simple for $n \geq 5$ (this is a major theorem, discussed below).
• $A_4$ is not simple: the Klein four-group $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$ is a proper nontrivial normal subgroup.

Why simple groups matter: atoms of group theory

Simple groups play the role for groups that prime numbers play for the integers. Every finite group can be “decomposed” into simple groups via composition series (see §15.7), and the Jordan-Holder theorem guarantees that this decomposition is essentially unique. Understanding all finite groups therefore reduces to:

1. Classifying all finite simple groups (completed in the 1980s—2004).
2. Understanding how simple groups can be assembled (the extension problem).

§15.5 Simplicity of $A_5$

Theorem 15.12. $A_5$ is simple.

Proof outline

We show that any normal subgroup $N \trianglelefteq A_5$ with $N \neq \{e\}$ must be all of $A_5$.

Key facts:

• $|A_5| = 60$.
• The conjugacy classes in $A_5$ have sizes: $1, 12, 12, 15, 20$ (corresponding to cycle types $e$, $(abcde)$, $(abced)$, $(ab)(cd)$, $(abc)$).
• A normal subgroup must be a union of conjugacy classes (since it is closed under conjugation) and must contain $e$.
• The order of $N$ must divide $60$.

Elimination: We need a union of conjugacy classes including $\{e\}$ whose total size divides $60$. The possibilities are:

$$1, \quad 1+12=13, \quad 1+12=13, \quad 1+15=16, \quad 1+20=21,$$ $$1+12+12=25, \quad 1+12+15=28, \quad 1+12+20=33,$$ $$1+15+20=36, \quad 1+12+12+15=40, \quad 1+12+12+20=45,$$ $$1+12+15+20=48, \quad 1+12+12+15+20=60.$$

Of these, the only ones dividing $60$ are $1$ and $60$.

Therefore $N = \{e\}$ or $N = A_5$, so $A_5$ is simple.

Remark. For $n \geq 5$, $A_n$ is simple. The proof for general $n$ uses the fact that $A_n$ is generated by $3$-cycles, and any normal subgroup containing a $3$-cycle must contain all $3$-cycles (by conjugation), hence must be all of $A_n$.

§15.6 The Center and the Commutator Subgroup

The center revisited

Definition 15.13. The center of $G$ is $Z(G) = \{z \in G : zg = gz \text{ for all } g \in G\}$.

We proved in §15.3 that $Z(G) \trianglelefteq G$. Note that $G$ is abelian if and only if $Z(G) = G$.

The commutator subgroup

Definition 15.14. For $a, b \in G$, the commutator of $a$ and $b$ is

The commutator subgroup (or derived subgroup) of $G$ is

the subgroup generated by all commutators.

Note: the set of all commutators is not always a subgroup (a product of two commutators need not itself be a commutator), so we must take the subgroup generated.

Theorem 15.15. $G' \trianglelefteq G$.

Proof

It suffices to show that $g[a,b]g^{-1}$ is a commutator for every $g, a, b \in G$, since $G'$ is generated by commutators.

Compute:

$$g[a,b]g^{-1} = g(aba^{-1}b^{-1})g^{-1} = (gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1})$$ $$= (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1} = [gag^{-1},\, gbg^{-1}].$$

So conjugating a commutator gives another commutator, which lies in $G'$. Since $G'$ is generated by commutators and conjugation sends generators to elements of $G'$, we conclude $gG'g^{-1} \subseteq G'$ for all $g$. Hence $G' \trianglelefteq G$.

Theorem 15.16. $G/G'$ is abelian.

Proof

Let $aG', bG' \in G/G'$. We need $(aG')(bG') = (bG')(aG')$, i.e., $abG' = baG'$.

Two cosets are equal if and only if their quotient lies in the subgroup, so it suffices to show

$$(ab)(ba)^{-1} \in G'.$$

But

$$(ab)(ba)^{-1} = aba^{-1}b^{-1} = [a,b],$$

and $[a,b] \in G'$ by definition of the commutator subgroup.

Therefore $abG' = baG'$, so

$$(aG')(bG') = (bG')(aG')$$

for all $a, b \in G$. Hence $G/G'$ is abelian.

Theorem 15.17. $G'$ is the smallest normal subgroup of $G$ with abelian quotient. That is, if $N \trianglelefteq G$ and $G/N$ is abelian, then $G' \subseteq N$.

Proof

Suppose $N \trianglelefteq G$ and $G/N$ is abelian. Then for all $a, b \in G$:

$$(aN)(bN) = (bN)(aN),$$

so $abN = baN$, hence $(ab)(ba)^{-1} \in N$, i.e., $aba^{-1}b^{-1} \in N$, i.e., $[a,b] \in N$.

Since every commutator lies in $N$ and $G'$ is generated by commutators, $G' \subseteq N$.

Example. For $S_3$: the commutators include $[(12),(123)] = (12)(123)(12)^{-1}(123)^{-1} = (12)(123)(12)(132) = (132)$. One checks that $S_3' = A_3 = \{e, (123), (132)\}$, and $S_3/S_3' \cong \mathbb{Z}_2$, which is abelian. For an abelian group $G$, $G' = \{e\}$.

Worked example 15.17a: the abelianization of $D_8$

Let

be the symmetry group of the square.

We want to compute the commutator subgroup $D_8'$ and the abelian quotient $D_8/D_8'$.

Start with a commutator:

Since $s^{-1}=s$ and $srs=r^{-1}$, we get

So

Now look at the quotient where all commutators are forced to disappear. In the abelianization, the relation

becomes

so

in the quotient. We already have $s^2=e$. Therefore the quotient is generated by the images of $r$ and $s$, both of order $2$, and it is abelian. So

Since $|D_8|=8$, the quotient has order $4$, so $|D_8'|=2$. But $D_8'$ already contains the nontrivial element $r^2$, hence

Therefore

Productive struggle: abelianization is not the same as quotienting by the center

Common wrong guess

To make a group abelian, quotient by its center.

Where it breaks. The center measures elements that already commute with everything. It does not measure all of the noncommutativity in the group. For $S_3$, the center is trivial:

$$Z(S_3)=\{e\}.$$

So

$$S_3/Z(S_3)\cong S_3,$$

which is still nonabelian.

Repaired method. The correct subgroup to quotient by is the commutator subgroup $G'$. The theorem says $G/G'$ is abelian and is the smallest such quotient. So the center and the derived subgroup answer different questions:

• $Z(G)$ asks which elements already commute with everything;
• $G'$ asks what must be killed to force the whole quotient to commute.

§15.7 The Second and Third Isomorphism Theorems

Theorem 15.18 (Second Isomorphism Theorem). Let $H \leq G$ and $N \trianglelefteq G$. Then $HN \leq G$, $N \trianglelefteq HN$, $H \cap N \trianglelefteq H$, and

Proof

$HN$ is a subgroup: Since $N$ is normal, for any $h \in H$ and $n \in N$, $hn = (hnh^{-1})h = n'h$ for some $n' \in N$. Hence $HN = NH$, and a product of two elements of $HN$ is:

$$(h_1 n_1)(h_2 n_2) = h_1 (n_1 h_2) n_2 = h_1 (h_2 n_1') n_2 = (h_1 h_2)(n_1' n_2) \in HN.$$

Closure under inverses is similar. So $HN \leq G$.

$N \trianglelefteq HN$: $N$ is normal in $G$, hence certainly in $HN \leq G$.

Define the map: Let $\phi: H \to HN/N$ be defined by $\phi(h) = hN$.

• Homomorphism: $\phi(h_1 h_2) = h_1 h_2 N = (h_1 N)(h_2 N) = \phi(h_1)\phi(h_2)$.
• Surjective: Every element of $HN/N$ has the form $hnN = hN$ (since $n \in N$), which is $\phi(h)$.
• Kernel: $\phi(h) = N \iff hN = N \iff h \in N$. Since $h \in H$, we get $h \in H \cap N$. So $\ker(\phi) = H \cap N$.

In particular, $H \cap N \trianglelefteq H$ (as the kernel of a homomorphism from $H$). By the FHT:

$$H/(H \cap N) \cong HN/N.$$

Example. Let $G = \mathbb{Z}_{12}$, $H = \langle\bar{4}\rangle = \{\bar{0}, \bar{4}, \bar{8}\}$, $N = \langle\bar{6}\rangle = \{\bar{0}, \bar{6}\}$.

Then $HN = \{\bar{0}, \bar{4}, \bar{8}, \bar{6}, \bar{10}, \bar{2}\} = \langle\bar{2}\rangle$, which has order $6$. And $H \cap N = \{\bar{0}\}$ (since $\bar{4}, \bar{8} \notin \{\bar{0}, \bar{6}\}$). The second isomorphism theorem gives:

Indeed, $|HN/N| = 6/2 = 3$, consistent.

Theorem 15.19 (Third Isomorphism Theorem). Let $N \subseteq M$ with both $N \trianglelefteq G$ and $M \trianglelefteq G$. Then $M/N \trianglelefteq G/N$ and

Proof

Define $\phi: G/N \to G/M$ by $\phi(gN) = gM$.

• Well-defined: If $gN = g'N$, then $g^{-1}g' \in N \subseteq M$, so $gM = g'M$.
• Homomorphism: $\phi(gN \cdot g'N) = \phi(gg'N) = gg'M = (gM)(g'M) = \phi(gN)\phi(g'N)$.
• Surjective: For any $gM \in G/M$, we have $\phi(gN) = gM$.
• Kernel: $\phi(gN) = M \iff gM = M \iff g \in M \iff gN \in M/N$. So $\ker(\phi) = M/N$.

In particular, $M/N \trianglelefteq G/N$. By the FHT:

$$(G/N)\big/(M/N) \cong G/M.$$

Example. Let $G = \mathbb{Z}$, $N = 12\mathbb{Z}$, $M = 4\mathbb{Z}$. Then $N \subseteq M$, both are normal in $\mathbb{Z}$. The theorem says:

The left side is $\mathbb{Z}_{12}/\langle\bar{4}\rangle$, which is a group of order $12/3 = 4$, and the right side is $\mathbb{Z}_4$. Indeed, $\mathbb{Z}_{12}/\langle\bar{4}\rangle \cong \mathbb{Z}_4$ as we computed in Chapter 14.

§15.8 Composition Series and the Jordan-Holder Theorem

Definition 15.20. A composition series for a group $G$ is a chain of subgroups

such that each composition factor $G_{i+1}/G_i$ is a simple group. (Each $G_i$ is normal in $G_{i+1}$, but not necessarily in $G$.)

Example: Composition series for $\mathbb{Z}_{12}$

The composition factors are:

• $\langle\bar{6}\rangle / \{0\} \cong \mathbb{Z}_2$
• $\langle\bar{2}\rangle / \langle\bar{6}\rangle \cong \mathbb{Z}_3$ (order $6/2 = 3$)
• $\mathbb{Z}_{12} / \langle\bar{2}\rangle \cong \mathbb{Z}_2$ (order $12/6 = 2$)

All factors are cyclic of prime order, hence simple. The composition factors (with multiplicity) are $\{\mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_2\}$.

An alternative composition series is $\{0\} \trianglelefteq \langle\bar{4}\rangle \trianglelefteq \langle\bar{2}\rangle \trianglelefteq \mathbb{Z}_{12}$ with factors $\mathbb{Z}_3, \mathbb{Z}_2, \mathbb{Z}_2$ — the same multiset, reordered.

Example: Composition series for $S_3$

Factors: $A_3/\{e\} \cong \mathbb{Z}_3$ and $S_3/A_3 \cong \mathbb{Z}_2$. Both simple. Composition factors: $\{\mathbb{Z}_2, \mathbb{Z}_3\}$.

Example: Composition series for $S_4$

where $V_4 = \{e, (12)(34), (13)(24), (14)(23)\}$.

This chain is not yet a composition series, because

is not simple: it has proper nontrivial subgroups. So we refine the chain by inserting one subgroup of order $2$ inside $V_4$.

A proper composition series for $S_4$ is:

Factors:

• $\langle(12)(34)\rangle / \{e\} \cong \mathbb{Z}_2$ (simple)
• $V_4 / \langle(12)(34)\rangle \cong \mathbb{Z}_2$ (simple)
• $A_4 / V_4 \cong \mathbb{Z}_3$ (simple, since $|A_4/V_4| = 12/4 = 3$)
• $S_4 / A_4 \cong \mathbb{Z}_2$ (simple)

Composition factors: $\{\mathbb{Z}_2, \mathbb{Z}_2, \mathbb{Z}_3, \mathbb{Z}_2\}$.

The Jordan-Holder Theorem

Theorem 15.21 (Jordan-Holder). If a finite group $G$ has a composition series, then any two composition series for $G$ have the same length and the same composition factors (up to permutation and isomorphism).

This theorem is stated without proof. It is the group-theoretic analogue of the uniqueness of prime factorization for integers. Just as every positive integer factors uniquely (up to order) into primes, every finite group decomposes uniquely (up to order) into simple composition factors.

Worked comparison: $\mathbb{Z}_6$ and $S_3$ have the same factors but are different groups

This is one of the most important sanity checks in the chapter.

For $\mathbb{Z}_6$, a composition series is

with factors

For $S_3$, a composition series is

with factors

So both groups have the same composition factors: