Group Theory - Jordan-Holder Decomposition

A group which has no proper normal subgroups is called a simple group.

Example: Cyclic groups of prime order are simple. Simple groups of composite order are "rare" according to the book.

A proper normal subgroup $A$ is called a maximum normal subgroup of $G$ if $A ◃ H ◃ G$ implies $H = G$ or $H = A$ . Note $A$ is a maximum invariant normal subgroup if and only if $G / A$ is a simple group, because $H / A$ is a normal subgroup of $G / A$ .

If $G$ is not simple, let $A$ a maximum normal subgroup in $G$ . Now if $A$ is not simple, let $A_{1}$ be a maximum normal subgroup. Continuing in this fashion we can construct a sequence, called a composition series as follows. $G ▹ A ▹ A_{1} ▹ . . . ▹ A_{r} ▹ {1}$ where $G / A, A / A_{1}, A_{1} / A_{2}, . . ., A_{r}$ are all simple nontrivial groups, which are called the composition quotient groups. The orders of the composition quotient groups are called the composition indices.

Jordan-Holder Theorem: In any two composition series for a group $G$ , the composition quotient groups are isomorphic in pairs, though may occur in different orders in the sequences.

Proof: Trivially the theorem is true if $∣ G ∣ = 2$ . Next assume the theorem has been proved for groups of order less than $∣ G ∣$ . If $G$ is simple then the theorem is again trivially true, otherwise let two composition series be $G ▹ A ▹ A_{1} ▹ . . . ▹ A_{r} ▹ {1}$ and $G ▹ B ▹ B_{1} ▹ . . . ▹ B_{s} ▹ {1}$

Then if $A = B$ , by inductive assumption the composition quotient groups $A / A_{1}, A_{1} / A_{2}, . . ., A_{r}$ and $G / B, B / B_{1}, B_{1} / B_{2}, . . ., B_{s}$ are isomorphic in pairs, and we have $G / A = G / B$ hence the theorem is true in this case.

Otherwise $A \neq B$ . Consider the group $A B$ . This contains $A$ and $B$ which are distinct and maximal in $G$ , thus we must have $A B = G$ . Let $D = A \cap B$ . By the first isomorphism theorem we have $G / A ≅ B / D$ and $G / B ≅ A / D$ . Note $G / A, G / B$ are simple, hence $B / D$ , $A / D$ are also simple which implies $D$ is in fact a maximum normal subgroup in $A$ and $B$ .

Now let $D ▹ D_{1} ▹ . . . ▹ D_{t} ▹ {1}$ be a composition series for $D$ .

Consider the quotient groups $G / A, A / D, D / D_{1}, . . ., D_{t}, {1}$ and $G / A, A / A_{1}, . . ., A_{r}, {1}$ By inductive assumption, the theorem is true for the group $A$ and hence the above sequences are isomorphic in pairs, and in fact $t = r$ .

Similarly $G / B, B / D, D / D_{1}, . . ., D_{t}, {1}$ is isomorphic in pairs to the sequence $G / B, B / B_{1},, . . ., B_{s}, {1}$ (and $s = r$ ).

But since the sequences $G / A, A / D, D / D_{1}, . . ., D_{t}, {1}$ and $G / B, B / D, D / D_{1}, . . ., D_{t}, {1}$ are clearly isomorphic in pairs, we have proved the theorem.

Example: The alternating group $A_{n}$ is a maximum normal subgroup of $S_{n}$ . We have already seen $A_{n}$ is normal in $S_{n}$ since it is of index 2. But the fact that it is of index 2 implies $S_{n} / A_{n}$ is simple and hence $A_{n}$ is maximal.

For $n = 3$ , we have the composition series $S_{3} ▹ A_{3} ▹ {1}$ since the composition indices are the primes $2,3$ .

For $n = 4$ , recall that the group $V = {1, (12) (34), (13) (24), (14) (23)}$ is normal in $A_{4}$ , and note every element in $V$ besides the identity generates a group of order $2$ of index $2$ (implying it is normal in $V$ ) thus we have the the composition series $S_{4} ▹ A_{4} ▹ V ▹ ⟨ (12) (34) ⟩ ▹ {1}$ with composition indices $2,3,2,2$ .

Example: Consider the cyclic group of order 6, and let $a$ be a generator. Then we have the composition series $⟨ a ⟩ ▹ ⟨ a^{2} ⟩ ▹ {1}$ with composition indices $2, 3$ . Note the composition quotient groups are isomorphic to those of $S_{3}$ , hence knowing the composition quotient groups is not enough to reconstruct the original group.

A group $G$ is said to be soluble if all the composition indices of $G$ are prime. For instance, all the groups in the above examples are soluble. Note a group $G$ is soluble if it contains a normal subgroup $H$ with both $G / H, H$ soluble. This is because given the series $H ▹ H_{1} ▹ . . . ▹ H_{r} ▹ {1}$ and $G / H ▹ G_{1} / H ▹ . . . ▹ G_{s} / H ▹ {1}$ with prime composition indices, we have $\frac{G_{i - 1} / H}{G_{i} / H} ≅ G_{i - 1} / G_{i}$ (where we set $G_{0} = G$ by applying the third isomorphism theorem. Hence we can construct the series with prime composition indices $G ▹ G_{1} ▹ . . . ▹ G_{s} ▹ H ▹ H_{1} ▹ . . . ▹ H_{r} ▹ {1}$

Lemma: If a normal subgroup $H$ of $A_{n}$ for $n \geq 3$ contains a cycle of degree $3$ then $H = A_{n}$ .

Proof: Without loss of generality let $(123) \in H$ . For $n = 3$ , $(123)$ generates $A_{3}$ and there is nothing to prove. For $n > 3$ , since $H$ is normal, it must also contain $s^{- 1} (123) s$ for any even permutation $s$ . Set $s = (32 k)$ for $k > 3$ . Then we have that $H$ contains $(1 k 2)$ , and hence also its square which is $(12 k)$ . Recall these cycles generate $A_{n}$ .

Theorem: $A_{n}$ is simple for $n > 4$ .

Proof: Suppose $H$ is a normal subgroup of $A_{n}$ . Suppose $h \in H$ is a permutation of the form $(a_{1} a_{2} . . . a_{m}) h'$ where $m > 3$ and $h'$ does not act on $a_{1}, . . ., a_{m}$ . Then the permutation $s = (a_{1} a_{2} a_{3})$ commutes with all the cycles of $h$ except the first, Now $s$ is even hence $h_{1} = s^{- 1} h s = (s^{- 1} (a_{1} . . . a_{m}) s h') \in H$ , thus $h_{1} h^{- 1} = (s^{- 1} a s) a^{- 1} = (a_{2} a_{3} a_{1} a_{4} . . . a_{m}) (a_{m} a_{m - 1} . . . a_{1}) = (a_{1} a_{3} a_{m}) \in H$ is contained in $H$ . Since this is a cycle of degree 3, by the above lemma we have $H = A_{n}$ . So if $H$ is to be a proper subgroup, its elements cannot contain cycles longer than 3.

Now suppose $H$ contains an element containing two 3-cycles. Without loss of generality, suppose $(123) (456) h' \in H$ where $h'$ does not act on $1,2,3,4,5,6$ . Set $s = (234)$ , so that it is an even permutation commuting with $h'$ . Then set $h_{1} = s^{- 1} h s = (134) (256) h' \in H$ , which gives $h_{1} h^{- 1} = (134) (256) (321) (654) = (12436) \in H$ which is a cycle of length greater than 3.

Now suppose $H$ contains an element containing exactly one 3-cycle, say $h = (123) h'$ , and $h'$ consists of 2-cycles implying $h'^{2} = 1$ . Then $h^{2} = (132)$ , so by the above lemma $H = A_{n}$ .

Lastly suppose $H$ consists only of permutations that are products of disjoint transpositions. For $n = 4$ this leads to the four-group $V$ in the above example. For $n > 4$ , suppose $h = (12) (34) h' \in H$ . Then set $s = (234)$ , and we have $h_{1} = s^{- 1} h s = (13) (42) h' \in H$ thus $h_{2} = h_{1} h^{- 1} = (13) (42) (12) (34) \in H$ Now take $t = (145)$ , and we have $h_{3} = t^{- 1} h_{2} t = (45) (23) \in H$ We conclude that $h_{3} h_{2}^{- 1} = (45) (23) (14) (23) = (45) (14) = (145) \in H$ , hence $H = A_{n}$ by the lemma.

Corollary: $A_{n}$ is the only subgroup of order $\frac{1}{2} n!$ in $S_{n}$ when $n > 4$ .

Proof: Any subgroup $H$ of index 2 is necessarily normal in $S_{n}$ , thus $D = A_{n} \cap H$ is normal in $A_{n}$ . By the Theorem we have $D = {1}$ or $D = {A_{n}}$ . Since $H$ contains more than one even permutation (because either half or all of a group of permutations are even) we must have $D = A_{n}$ , implying $H = A_{n}$ .

It can be easily verified that the statement of the corollary is also true for $n \leq 4$

Corollary: $S_{n}$ is not soluble for $n > 4$ .

Proof: By the theorem, the composition series for $S_{n}$ is $S_{n} ▹ A_{n} ▹ {1}$ and its composition indices are $2, \frac{1}{2} n!$ , the latter of which is not prime.