## 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 $A \triangleleft H \triangleleft G$ 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.

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

and

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 $A \ne B$. 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 $G/A \cong B/D$ and $G/B \cong A / D$. 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 $D\triangleright D_1 \triangleright ... \triangleright D_t \triangleright \{1\}$ be a composition series for $D$.

Consider the quotient groups

and

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

is isomorphic in pairs to the sequence

(and $s = r$).

But since the sequences

and

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

since the composition indices are the primes $2,3$.

For $n=4$, recall that the group $V=\{1,(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)\}$ 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

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

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

and

with prime composition indices, we have $\frac{G_{i-1} /H}{G_i / H} \cong G_{i-1} / G_i$ (where we set $G_0 = G$ by applying the third isomorphism theorem. Hence we can construct the series with prime composition indices

**Lemma:** If a normal subgroup $H$ of $A_n$ for $n \ge 3$
contains a cycle of degree $3$ then $H = A_n$.

**Proof:** Without loss of generality let $(1 2 3) \in H$.
For $n = 3$, $(1 2 3)$ generates $A_3$ and there is nothing to prove.
For $n \gt 3$, since $H$ is normal, it must also
contain $s^{-1} (1 2 3) s$ for any even permutation $s$. Set
$s = (3 2 k)$ for $k \gt 3$.
Then we have that $H$ contains $(1 k 2)$, and hence also
its square which is $(1 2 k)$.
Recall these cycles generate $A_n$.

**Theorem:** $A_n$ is simple for $n \gt 4$.

**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 $m \gt 3$ 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

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 $(1 2 3)(4 5 6)h' \in H$ where $h'$ does not act on $1,2,3,4,5,6$. Set $s = (2 3 4)$, so that it is an even permutation commuting with $h'$. Then set $h_1 = s^{-1} h s = (1 3 4)(2 5 6)h' \in H$, which gives

which is a cycle of length greater than 3.

Now suppose $H$ contains an element containing exactly one 3-cycle, say $h = (1 2 3) h'$, and $h'$ consists of 2-cycles implying $h'^2 = 1$. Then $h^2 = (1 3 2)$, 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 $n\gt 4$, suppose $h = (1 2)(3 4)h' \in H$. Then set $s = (2 3 4)$, and we have

thus

Now take $t = (1 4 5)$, and we have

We conclude that $h_3 h_2^{-1} = (4 5)(2 3)(1 4)( 2 3) = (4 5)(1 4) = (1 4 5) \in H$, hence $H = A_n$ by the lemma.

**Corollary:** $A_n$ is the only subgroup of order $\frac{1}{2}n!$
in $S_n$ when $n \gt 4$.

**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 $n \le 4$

**Corollary:** $S_n$ is not soluble for $n \gt 4$.

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