We need a few remarks before we jump into the proof. The first one is that if H and K are maximal normal subgroups of a group G which are distinct then HK = G. This is very easy to show. By definition H and K are proper normal subgroups of G which is not contained in some other proper normal subgroup of G. By construction HK is the smallest group containing both H and K. Since both H and K are normal it follows that HK is also normal. Thus, HK is a normal subgroup of G containing both H and K. In addition if H and K are distinct then HK would contain both H and K properly. This forces HK = G because the only normal subgroup which can contain H and K properly must be the full group G. The second one is that every

**finite**group has a composition series, go back here if you would like to see the definition of this term. Thus far all groups we dealt with we have assumed they have a composition series. In the case of finite group we can prove it must have a composition series. The proof of this is also very simple. Recall our observation that H is a maximal normal subgroup of G if and only if G/H is a (non-trivial) simple group. For a given finite group G consider all of its normal subgroups, there are finitely many, so we can choose a maximal one. Now look at this maximal one and find its maximal one. Keep on going until you reach no more, this would be a composition series.

**Theorem:**Any two composition series for a finite group must be isomorphic.

**Proof:**We need to prove two statements. First, if $$\{H_i\}_{i=0}^n$$ and $$\{K_i\}_{i=0}^m$$ are composition series then $$n=m$$ and furthermore there is a one-to-one correspondence between the factor groups $$H_{i+1}/H_i$$ and $$K_{i+1}/K_i$$ which give an isomorphism. We prove this by induction on |G|. Notice that if the order of G was 1,2,3,4,5, the few basic cases, then we can easily see that the theorem is true. We would like to prove this theorem for larger orders. So we will assume it is true for all smaller orders than |G| and prove it for |G|. We will change our notation slightly. Instead, of $$H_0\subset ... \subset H_n$$ being a composition series with $$H_0$$ being the trivial group and $$H_n=G$$ we will write the indices in reverse, $$H_n\subset ... \subset H_0$$. Likewise, we will also write $$K_m \subset ... \subset K_0$$ where $$K_m$$ is trivial and $$K_0=G$$. Consider $$H_1$$ and $$K_1$$, by our new notation those are our maximal normal subgroups of G. There are two possibilities, either they are both equal or they are distinct. If they are both equal then the proof is finished by induction. Because $$\{H_i\}_{i=n}^1$$ is a composition series for $$H_1$$ and $$\{K_i\}_{i=m}^1$$ is a compsotition series for $$K_1$$. Since $$H_1=K_1$$ these are two composition series for a group or order less than |G|. Thus, it follows that $$n=m$$ and the factor groups can be arranged to give rise to isomorphisms. The harder case is when these two groups are distinct. Construct $$N_1=H_1\cap K_1$$. It follows that $$N_1$$ is a normal subgroup of G. Let $$\{N_i\}_{i=r}^1$$ be a composition series for $$N_1$$. Now consider the following subnormal series $$N_r\subset ... \subset N_1 \subset H_1$$. We claim this is a composition series for $$H_1$$. To prove this we only need to show that $$H_1/N_1$$ is a simple group. By the second isomorphism theorem $$H_1K_1/K_1 \simeq H_1/(H_1\cap K_1)$$. Thus, $$G/K_1$$ is isomorphism to $$H_1/N_1$$, but $$G/K_1$$ is simple, so we do indeed have a composition series. But $$H_n \subset ... \subset H_1$$ is also a composition series for $$H_1$$. By induction it follows that r+1=n. Furthermore, the two factor groups can be rearranged to have an isomorphism. By repeating the similar argument with $$K_1$$ we see that $$N_r\subset ... \subset N_1 \subset K_1$$ is a composition series and $$K_m\subset ... \subset K_1$$ is a composition series. Thus, r+1=m also, and the factor groups can be rearranged to give rise to isomorphisms. It follows from all of this that n=m. Now we will prove that the two composition series for G have an isomorphic pairing of their factor groups. For this consider $$N_r\subset ... \subset N_1 \subset H_1$$ and $$H_n\subset ... \subset H_1$$. As we explained the factor groups must pair isomorphically. Thus, there is an isomorphic correspondence betweeen $$N_i/N_{i+1}$$ and $$H_i/H_{i+1}$$ except for one factor group which happens to be isomorphic to $$H_1/N_1$$. Likewise, by considering $$N_r \subset ... \subset N_1 \subset K_1$$ and $$K_n \subset ... \subset K_1$$ we again see there is an isomorphic correspondence between $$N_i/N_{i+1}$$ and $$K_i/K_{i+1}$$ except for one factor group which happens to be isomorphic to $$K_1/N_1$$. Putting this together we see that in the composition series $$H_n \subset ... \subset H_1 \subset G$$ and $$K_n \subset ... \subset K_1 \subset G$$ we see there is a one-to-one isomorphic correspondence between all groups $$H_i/H_{i+1}$$ but one which happens to be isomorphic to $$H_1/N_1$$ and the groups $$K_i/K_{i+1}$$ but one which happens to be isomorphic to $$K_1/N_1$$. Now take the group which happens to be isomorphic to $$H_1/N_1$$ and pair it with $$G/K_1$$ which is isomorphic by the second isomorphism theorem. Likewise take the group which happens to be isomorphic to $$K_1/N_1$$ and pair it with $$G/H_1$$ which is isomorphic by the second isomorphism theorem. Thus, we have obtained a one-to-one isomorphic correspondence between $$\{ H_i\}_{i=n}^0$$ and $$\{ K_i\}_{i=n}^0$$. Q.E.D.

The Jordan Holder theorem has a lot of interesting applications. One of which is the concept of a "solvable group" (or as English mathematicians in contrast to American mathematicians like to write "soluble groups"). Solvable groups comes up when we wish to determine if a polynomial is solvable in radicals (over some field). The famous result from Galois that there are degree five polynomials not solvable in radicals is based on the idea that $$S_n$$ is not a solvable for $$n\geq 5$$. We will state a definition.

**Definition:**A (finite) group G is solvable iff we can write a subnormal series $$\{N_i\}_{i=0}^n$$ such that $$N_{i+1}/N_i$$ is abelian.

There is an equivalent definition. We can insist for $$N_{i+1}/N_i$$ to be prime cyclic groups (and hence simple). That is easy to show. If $$N_0\subseteq ... \subseteq N_n$$ is a solvable series then we can start inserting factors in between adjacent groups. If $$N_{i+1}/N_i$$ is not simple then we can find a group $$N_i \subset K \subset N_{i+1}$$ (by taking inverse images under the natural projection map). Since the group is finite (well we are assuming that for simplicity sake, pun intended) it follows that eventually we reach a point when we cannot insert any other groups in between. Thus, we would have refined a solvable series into one where all factor groups are simple. Since all the factor groups stay abelian it follows that the factor groups must be prime cyclic groups, because those are the only non-trivial abelian groups which are simple. Thus, we can define a (finite) group G to be solvable iff we can write a subnormal series $$\{N_i\}_{i=0}^n$$ such that $$N_{i+1}/N_i$$ are prime cyclic groups.

By this useful equivalent definition we can formulate another equivalent definition. A (finite) group G is solvable if and only if it has a composition series with abelian factor groups. Here is the important point. By the Jordan Holder theorem it follows that if one composition series of G has abelian factor groups then it follows that all composition series of G have abelian factor groups. Thus, if there is a composition series that has a factor group which is not abelian then the group cannot be solvable. This simple (pun intended) test can be used to show that a group is not solvable. The standard example is the next theorem.

**Theorem:**For $$n\geq 5$$ the symmetric group $$S_n$$ is not solvable.

**Proof:**Recall that $$A_n$$ is simple for $$n\geq 5$$. Thus, $$\{1\}\subset A_n \subset S_n$$ is a composition series where $$A_n/\{1\} \simeq A_n$$ is not abelian. Thus, by the previous comments $$S_n$$ cannot be solvable.

As an excerise try proving that $$A_n$$ is the only proper normal non-trivial subgroup of $$S_n$$ for $$n\geq 5$$. This proof would be much more difficult if we did not have the group theoretic machinery of the Jordan Holder theorem.

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