**Definition:**Let G be a group, let $$H_i$$ be subgroups for $$0\leq i \leq n$$ (for some integer n). So that $$H_0\subseteq H_1\subseteq ... \subseteq H_n$$ with $$H_0$$ being the trivial subgroup and $$H_n=G$$. We call this group series a "subnormal series" iff $$H_i$$ is a normal subgroup of $$H_{i+1}$$. If in addition we have proper containment $$\subset$$ rather than just containment $$\subseteq$$ we will refer to such a subnormal series as "reduced".

**Definition:**Let G be a group with $$\{H_i\}_{i=0}^n$$ as a subnormal series and let $$\{K_i\}_{i=0}^m$$ be another subnormal series. We say that $$K_i$$ is "finer" (or a "refinement" of $$H_i$$) than $$H_i$$ if the all $$H_i$$ are contained in the sequence of the $$K_i$$.

Non-reduced series do not add any new information, we can cancel out all extra groups on the list. We can therefore say in our terminology that any subnormal series is always a refinement of some reduced subnormal series. Here is an example. Notice that $$ 0\mathbb{Z} \subset 4\mathbb{Z} \subset \mathbb{Z}$$ can be refined to $$ 0\mathbb{Z} \subset 4\mathbb{Z} \subset 2\mathbb{Z} \subset \mathbb{Z}$$. A way to think of a refinement is by adding intermediate groups in between the already existing ones. Of course, sometimes it is not possible to add any (distinct) intermediate groups in between. Suppose we have $$0\mathbb{Z} \subset p^2 \mathbb{Z} \subset p\mathbb{Z} \subset \mathbb{Z}$$ then it is not possible to add any distinct subgroup in between those two if p is a prime. The series definitely has a refinement we can add $$p^3\mathbb{Z}$$ in between $$0\mathbb{Z}$$ and $$p^2\mathbb{Z}$$. But we cannot add a subgroup in between $$p$$ and $$p^2$$.

What we are going to be going is considering the factor group $$H_{i+1}/H_i$$. Sometimes two different looking series can turn out to the same by considering their factor groups. For that we will give a definition.

**Definition:**Two subnormal series $$\{H_i\}_{i=0}^n$$ and $$\{K_i\}_{i=0}^n$$ are said to be "isomorphic" iff there is a one-to-one correspondence between the factor groups $$H_{i+1}/H_i$$ and $$K_{i+1}/K_i$$ so that the correspondence is an isomorphism of groups.

For example, consider $$0\mathbb{Z}_{10} \subset 2\mathbb{Z}_{10} \subset \mathbb{Z}_{10}$$ and $$0\mathbb{Z}_{10}\subset 5\mathbb{Z}_{10}\subset \mathbb{Z}_{10}$$. Notice that $$2\mathbb{Z}_{10}/0\mathbb{Z}_{10}\simeq \mathbb{Z}_5$$ and $$\mathbb{Z}_{10}/2\mathbb{Z}_{10} \simeq \mathbb{Z}_2$$. But $$5\mathbb{Z}_{10}/ 0\mathbb{Z}_{10}\simeq \mathbb{Z}_2$$ and $$\mathbb{Z}_{10}/ 5\mathbb{Z}_{10} \simeq \mathbb{Z}_5$$. So we clearly have an isomorphism between these two subnormal series.

Now we will get to the theorem.

**Theorem:**Every two subnormal series of groups have isomorphic refinements. That is to say if $$ \{ H_i\}_{i=0}^n $$ and $$ \{K_i\}_{i=0}^m $$ are a subnormal series of G then we can find series $$\{ \hat H_i\}_{i=0}^N $$ and $$ \{\hat K_i\}_{i=0}^N$$ so that $$\hat H_i$$ is a refinement of $$H_i$$ and $$\hat K_i$$ is a refinement of $$K_i$$, while the additional property that $$\{\hat H_i\} \simeq \{ \hat K_i\}$$.

**Proof:**The proof is very nice. For every i notice that we have $$H_i = H_i(H_{i+1}\cap K_0)$$ and $$H_{i+1}=H_i(H_{i+1}\cap K_m)$$. Notice that we also have the following inclusions:

$$H_i(H_{i+1}\cap K_0)\subseteq H_i(H_{i+1}\cap K_1)\subseteq ... \subseteq H_i(H_{i+1} \cap K_m)$$

Let us define $$H_{ij} = H_i(H_{i+1}\cap K_j)$$ for $$0\leq i\leq n-1$$ and $$0\leq j \leq m$$. By the above remarks we have $$H_{i,0}=H_i$$ and $$H_{i,m}=H_{i+1,0}=H_{i+1}$$.

Now consider, the following inclusions:

$$H_{00}\subseteq H_{01}\subseteq H_{02}\subseteq ... \subseteq H_{0,m-1}\subseteq H_{0m}=H_{10}\subseteq$$

$$\subseteq ~ ~ ~ ~ ~ ~ H_{11}\subseteq H_{12}\subseteq ... \subseteq H_{1,m-1}\subseteq H_{1m} = H_{20}\subseteq$$

......

$$\subseteq ~ H_{n-1,1}\subseteq H_{n-1,2}\subseteq ... \subseteq H_{n-1,m-1}\subseteq H_{n-1,m} = G$$

This long inclusion sequence definitely contains out original $$H_i$$ so it is a refinement. Now we will define a similar one for the $$K_i$$ subnormal series by defining $$K_{ij} = K_i(K_{i+1}\cap H_j)$$ for $$0\leq i\leq m-1$$ and $$0\leq j \leq n$$. And now we can define a similar long inclusion sequence for the $$K_{ij}$$. I will not bother writing out because I think it is clear. From the diagrams it is also clear that once we take factor group $$H_{i,j+1}/H_{i,j}$$ then we are left with a rectangle having m coloumns and n rows. While the $$K_{i,j+1}/K_{i,j}$$ is a rectangle having n columns and m rows. Furthermore, we see most importantly that $$H_{i+1,j}/H_{i,j}\simeq K_{j+1,i}/K_{j,i}$$. Because $$H_{i}$$ is normal in $$H_{i+1}$$ and $$K_{j}$$ is normal in $$K_{j+1}$$ so it follows by the fourth isomorphism theorem that:

$$_{H_i(H_{i+1}\cap K_{j+1})/H_i(H_{i+1}\cap K_j)\simeq K_j(H_{i+1}\cap K_{j+1})/K_j(H_i\cap K_{j+1})}$$. This is what we wanted, we have established that the factor groups are isomorphic. Q.E.D.

The Schreier theorem gives rise to an important corollary in group theory known as the Jordan-Holder theorem. Before we state the theorem we need an additional definition.

**Definition:**A reduced subnormal series $$H_i$$ is said to be a "composition series" when $$H_{i+1}/H_i$$ is a simple group.

Here is a useful remark. Let G be a group with normal subgroup N. There is a one-to-one correspondence between the subgroups of G which contain N and the subgroups of G/N. Furthermore, this correspondence is still preserved when we limit our selves to normal subgroups of G which contain N and normal subgroup of G/N. The proof is very simple. Notice that the natural homomorphism $$\pi:G\to G/N$$ provides all the correspondence we need. If K is a (normal) subgroup then $$\pi (K) = K/N$$ is a (normal) subgroup of G/N. Therefore, it follows N is a maximal normal subgroup of G if and only if G/N is simple. This means that an equivalent way to formulate a composition series is by saying that $$H_i$$ is a maximal normal subgroup of $$H_{i+1}$$.

**Lemma:**If $$\{K_i\}$$ and $$\{K_i^|\}$$ are isomorphic series which are refinements of the reduced series $$\{ H_i\}$$ and $$\{H_i^|\}$$ respectively then it follows that $$\{H_i\}\simeq \{H_i^|\}$$ also.

**Proof:**Instead of writing out a full formal proof here is an easy way to see why this must be true. On the left side of paper draw a coloumn of factor groups $$K_{i+1}/K_i$$. On the right side of paper draw a coloumn of factor groups $$K^|_{i+1}/K^|_i$$. Now draw lines connecting isomorphic factor groups. (This will look similar to be bipartite graph, but not quite). The zero groups on the left are precisely the ones where the $$K_i=K_{i+1}$$, likewise on the right. Because of the isomorphism the number of zero groups on the left will equal the number of zero groups on the right. Erase the zero lines which connect those groups. The groups we are left with now form a reduced series and furthermore the remaining lines provide the isomorphism that we desired. Q.E.D.

Here is the very surprising result. One of my favorite theorems. Totally unexpected. This is the Jordan-Holder theorem.

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

**Proof:**The proof is incredibly simple once with have the Schreier theorem avaliable. By the Schreier theorem any two composition series have isomorphic refinements. However, by the lemma, the reduced series will themselves be isomorphic. But the reduced series are the composition series themselves! This is because a composition series cannot be refined into a proper reduced series by our remarks above. Thus, it follows that the composition series must be isomorphic.

This is a surprising result and an important result from group theory. We will give another additional proof of this theorem using the second isomorphism theorem rather than the fourth. We will also prove it directly without invoking the use of the Schreier theorem.

## No comments:

## Post a Comment