# Jordan–Hölder Theorem

The **Jordan–Hölder theorem** is a theorem about composition series of finite groups. A **composition series** is a chain of subgroups
$1 = H_0 \triangleleft H_1 \triangleleft H_2 \triangleleft \cdots \triangleleft H_{k-1} \triangleleft H_k = G,$
where $H_i$ is a maximal proper normal subgroup of $H_{i+1}.$ By the third isomorphism theorem, this is equivalent to the statement that the quotient $H_{i+1}/H_i$ is a simple group. This quotient is called a **composition factor**.

It is not hard to show that every finite group $G$ has a composition series. The Jordan–Hölder theorem states that any two composition series of the same group have the same length and the same composition factors (up to permutation).

The cyclic group ${\mathbb Z}_6$ has two composition series: $1 \triangleleft {\mathbb Z}_3 \triangleleft {\mathbb Z}_6 \, \, \text{and} \, \, 1 \triangleleft {\mathbb Z}_2 \triangleleft {\mathbb Z}_6.$ Here, ${\mathbb Z}_3$ is isomorphic to the subgroup $\{0,2,4\}$ of ${\mathbb Z}_6,$ and ${\mathbb Z}_2$ is isomorphic to the subgroup $\{0,3\}.$ Note that both composition series have the same length, and both have the same composition factors, but in a different order: ${\mathbb Z}_6/{\mathbb Z}_3 \cong {\mathbb Z}_2$ and ${\mathbb Z}_6/{\mathbb Z}_2 \cong {\mathbb Z}_3.$

## Background

Here is a discussion of two assertions made in the introduction: that every finite group has a composition series, and that $H_i$ being a maximal normal subgroup of $H_{i+1}$ is equivalent to saying that $H_{i+1}/H_i$ is simple.

It is clear that every finite group has a composition series. To show that a maximal normal subgroup exists, the trivial subgroup is normal in every group, and if it is not maximal, there must be a larger normal subgroup containing it. Continue looking for larger normal subgroups until finding one that is maximal. The process is finite because the group is finite. The maximal normal subgroup is $H_{k-1}.$ Run the same process on subgroups of $H_{k-1}.$

The third isomorphism theorem states that normal subgroups of $G/N$ are in one-to-one correspondence with normal subgroups of $G$ containing $N,$ via the natural correspondence coming from the standard homomorphism $\pi \colon G \to G/N.$ So $H_i$ is maximal in $H_{i+1}$ if and only if there is no normal subgroup strictly containing $H_i$ and strictly contained in $H_{i+1},$ which corresponds to there being no normal subgroup in $H_{i+1}/H_i$ which is strictly bigger than $\{1\}$ and strictly smaller than the whole quotient. This is the same as saying that $H_{i+1}/H_i$ is simple.

**Warning**: Normality is not a transitive relation. That is, if $K$ is normal in $H$ and $H$ is normal in $G,$ it is not necessarily true that $K$ is normal in $G.$ Each of the subgroups in a composition series is normal in the succeeding one, but not necessarily normal in $G$ itself. For example, $S_4$ has a composition series

$1 \triangleleft {\mathbb Z}_2 \triangleleft V_4 \triangleleft A_4 \triangleleft S_4$
(see the simple group wiki for a derivation), but ${\mathbb Z}_2,$ the subgroup generated by a double transposition, is not normal in $S_4.$

## Formal Statement

Let $G$ be a finite group. Consider two composition series

$\begin{aligned} 1 = H_0 \triangleleft H_1 \triangleleft H_2 \triangleleft \cdots \triangleleft H_{k-1} \triangleleft H_k &= G \\ 1 = K_0 \triangleleft K_1 \triangleleft K_2 \triangleleft \cdots \triangleleft K_{\ell-1} \triangleleft K_\ell &= G. \end{aligned}$

Then $k=\ell$ and the list of composition factors is unique up to permutation; that is, the lists $\{H_{i+1}/H_i\}$ and $\{K_{j+1}/K_j\}$ are the same, after rearranging one of the lists suitably.

## Applications

**Unique factorization:**The Jordan–Hölder theorem can be viewed as a generalization of the fundamental theorem of arithmetic that every integer can be factored as a product of prime numbers, essentially uniquely (up to permutation of the factors).

In fact, it is not hard to show that the fundamental theorem of arithmetic follows from Jordan–Hölder: consider the cyclic group ${\mathbb Z}_n.$ Every subgroup is normal, and a maximal subgroup in a cyclic group is precisely one which has prime index. (This is because the quotient, being abelian, is simple if and only if it contains no nontrivial proper subgroups, which is only true if the quotient has prime order.) So a composition series for ${\mathbb Z}_n$ must have the form $1 \triangleleft {\mathbb Z}_{p_1} \triangleleft {\mathbb Z}_{p_1p_2} \triangleleft \cdots \triangleleft {\mathbb Z}_{p_1p_2\cdots p_k} = {\mathbb Z}_n,$ where the $p_i$ are primes. Since ${\mathbb Z}_n$ has a composition series, $n$ can be factored as a product of primes, and by Jordan–Hölder, a different factorization of $n$ leads to a different composition series $1 \triangleleft {\mathbb Z}_{q_1} \triangleleft {\mathbb Z}_{q_1q_2} \triangleleft \cdots \triangleleft {\mathbb Z}_{q_1q_2\cdots q_\ell} = {\mathbb Z}_n,$ which must have the same composition factors up to permutation $($and $k=\ell).$ But the orders of the composition factors are just the primes $p_i\ (\text{and } q_j),$ so these lists are the same up to permutation, which is precisely the statement of the fundamental theorem of arithmetic.

**Normal subgroups of the symmetric group:**For $n\ge 5,$ the alternating group $A_n$ is simple, so there is a composition series $1 \triangleleft A_n \triangleleft S_n.$ Since the composition factor sizes are $2$ and $\frac{n!}{2},$ this implies that the only other possible nontrivial normal subgroup of $S_n$ has order $2,$ which can easily be ruled out by a simple analysis of the nontrivial element of the subgroup. So, for $n \ge 5,$ the only nontrivial proper normal subgroup of $S_n$ is $A_n.$ This is related to solvability of polynomial equations by radicals.**Algebraic geometry:**A slight generalization of Jordan–Hölder (to modules) allows for an elegant and robust definition of the multiplicity of the intersection of two algebraic curves. This is used in Bezout's theorem, which predicts the number of intersection points of two projective curves. The details of the definition are beyond the scope of this wiki, but the idea is that the intersection multiplicity is defined as the*length*of a certain module, which is defined to be the length of its composition series. The composition series are not unique, but they all have the same number of terms, thanks to Jordan–Hölder.

## Proof of the Theorem

This proof is fairly technical. It will help to compare with the proof of the fundamental theorem of arithmetic, and to understand the second isomorphism theorem.

As with the fundamental theorem of arithmetic, the proof proceeds by induction, on $|G|.$ The base case $|G|=1$ is trivial. Now suppose the theorem has been proven for all groups strictly smaller than $G.$

Take two composition series $(H_1,H_2,\ldots,H_k)$ and $(K_1,K_2,\ldots,K_\ell)$ for $G.$ The theorem is true for $H = H_{k-1}$ and $K = K_{\ell-1}.$ If $H=K,$ then we are done, as the composition series must be rearrangements of each other. If $H \ne K,$ let $L = H \cap K.$ Then $L$ has a composition series consisting of groups $L_j,$ by the inductive hypothesis. Then there are two composition series for $H,$ the one involving the $H_i$ and the following one: $1 \triangleleft L_1 \triangleleft L_2 \triangleleft \cdots \triangleleft L_{t-1} \triangleleft L \triangleleft H.$ $(L=H \cap K$ is a maximal subgroup of $H$ because $H/(H\cap K) \cong G/K,$ by the second isomorphism theorem.$)$

By induction, this composition series must be a rearrangement of the other one: $(H_1/H_0,H_2/H_1,\ldots,H/H_{k-2}) \sim (L_1/L_0,L_2/L_1,\ldots,L/L_{t-1},H/L),$ where $\sim$ means "is the same up to permutation." Note that the lengths being the same implies that $t+1=k.$

Similarly we get two composition series for $K,$ using the same $L_i$ series for the second one. That is, $(K_1/K_0,K_2/K_1,\ldots,K/K_{\ell-2}) \sim (L_1/L_0,L_2/L_1,\ldots,L/L_{t-1},K/L).$ So $t+1=\ell.$ This proves that $k=\ell.$

Now append $G/H$ to the first pair of lists, and append $G/K$ to the second pair of lists. This gives $\begin{aligned} (H_1/H_0,H_2/H_1,\ldots,H/H_{k-2},G/H) &\sim (L_1/L_0,L_2/L_1,\ldots,L/L_{t-1},H/L,G/H)\\ (L_1/L_0,L_2/L_1,\ldots,L/L_{t-1},K/L,G/K) &\sim (K_1/K_0,K_2/K_1,\ldots,K/K_{\ell-2},G/K). \end{aligned}$ We want that the outer two lists are the same up to permutation. The inner two lists are the same except for the last two entries. But $(H/L,G/H)$ and $(K/L,G/K)$ are the same as $(G/K,G/H)$ and $(G/H,G/K),$ again by the second isomorphism theorem. So the inner two lists are the same up to permutation (a transposition of the last two factors), and the result follows. $_\square$

**Cite as:**Jordan–Hölder Theorem.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/jordan-holder/