---- > [!definition] Definition. ([[composition series]]) > A **composition series** for a [[group]] $G$ is a [[normal series]] in which each successive [[quotient group|quotient]] is nontrivial and [[simple group|simple]]. - [ ] include definition modules too > [!basicexample] **$S_{3}$.** $S_{3}$ has a single [[normal subgroup]], $H:=\{ e, \tau, \tau^{2} \}$. Since $S_{3} / H$ [[order of quotient group is quotient of orders|has order 2]], it is [[group isomorphism|isomorphic]] to $C_{2}$ and thus [[simple group|simple]]. Clearly $H / \{ e \}=H \cong C_{3}$ is [[simple group|simple]]. So only [[composition series]] is $S_{3} \trianglelefteq H \trianglelefteq (e)$. The simple factors are $C_{2}$ and $C_{3}$. **$D_{4}$.** Be careful! The [[subgroup]] [[lattice (groups)]] of $D_{4}$ is ![[CleanShot 2023-09-03 at [email protected]]] ([[dihedral group|source]]). Notice that each [[subgroup]] of order 4 has [[index of a subgroup|index]] $2$ in $D_{4}$ and is hence [[subgroups of index 2 are normal|is normal in]] $D_{4}$. Furthermore, each [[subgroup]] of order $2$ has index $2$ in each [[subgroup]] of order $4$ containing it, and is hence normal in that [[group]]. Finally, $(e)$ is a [[normal subgroup]] of any group. Therefore any path on four nodes from $(e)$ to $D_{4}$ in the lattice corresponds to a [[normal series]]. There are $7$ total. Are they all [[composition series]]? Yes, because the [[quotient group|quotient]] at each step has two elements and is hence [[group isomorphism|isomorphic]] to $C_{2}$. So the simple factors are $C_{2}$, $C_{2}$, and $C_{2}$. **$Q_{8}$.** ![[CleanShot 2023-09-17 at 18.07.55.jpg]] Recall: [[quaternion group is nonabelian, but all its subgroups are normal]]. So any [[normal series]] of $Q$ is obtained by taking a $4$-node path from $\{ \b 1 \}$ to $Q$. There are $3$ such paths, hence $3$ [[normal series]]. Are they all [[composition series]]? Yes, because the order $4$ [[subgroup]]s have [[subgroups of index 2 are normal|index 2]] in $Q$ and the order $2$ [[subgroup]] has [[subgroups of index 2 are normal|index 2]] in any order $4$ [[subgroup]]. **$C_{3} \times C_{3}$.** Here is the lattice: ![[CleanShot 2023-10-16 at 20.41.46.jpg|400]] $C_{3} \times C_{3}$ is an [[abelian group]], so all the [[subgroup]]s are [[normal subgroup|normal]]. Each [[quotient group]] between levels is [[simple group|simple]] because it has [[order of a group|order 3]]. Thus any $3$-node path from $(e)$ to $C_{3} \times C_{3}$ is a [[composition series]]. **$C_{9}$**. Via [[uniqueness of subgroups of finite cyclic groups]], $C_{9}$ has a unique proper [[subgroup]] of order $3$ which is [[normal subgroup|normal]] since $C_{9}$ is [[abelian group|abelian]]. There is just the one [[composition series]]: $C_{9} \trianglerighteq C_{3} \trianglerighteq (e)$. **$S_{5}$**. Using [[A5 is the unique simple group of order 60]], we know that *a* [[composition series]] is $S_{5} \trianglerighteq A_{5} \trianglerighteq (e)$. We claim it is the only [[composition series]]. ![[CleanShot 2023-10-16 at 20.47.11.jpg|500]] One reason for this is that if not, then $S_{5}$ has another [[normal subgroup]] $N$. That $N \cap A_{5} \trianglelefteq S_{5}$ and $A_{5}$ simple implies $N \cap A_{5}=A_{5}$ or $N \cap A_{5}=(e)$. But the second is not possible since $N$ needs to contain nontrivial [[parity of a permutation|even permutation]]s to be a [[subgroup]] (since $\text{odd}\circ \text{odd}=\text{even}$). So $N \cap A_{5}=A_{5}$. By assumption $N$ is proper and $A_{5}$ has the highest possible order of a [[normal subgroup]] of $S_{5}$, thus in fact $N=A_{5}$. $A_{4}$. In [[alternating group#^528873|this example]] we found the elements of $A_{4}$ are ^c1fedb - 8 $3$-cycles: $(123), (124), (132), (134), (142), (143), (234), (243)$ - 3 disjoint products of $2$-cycles in $S_{4}$: $(12)(34), (13)(24), (14)(23)$ - $(e)$. Let $N$ be a [[normal subgroup]] of $A_{4}$ such that $A_{4} / N$ is [[simple group|simple]]. Then the [[index of a subgroup|index]] of $N$ in $A_{4}$ is either $2$ or $3$. ![[CleanShot 2023-10-25 at [email protected]]] If $[A_{4} : N]=2$, then $|N|=6$. Set $K:=\{ e, (12)(34), (13)(24), (14)(23) \}$. The $2$-cycles each square to the identity, [[Klein 4-group|hence]] $K \cong C_{2} \times C_{2}$. The elements of order 2, as disjoint-$2$-cycle-products, are all [[conjugate]] in $A_{4}$ (can [[conjugation in the symmetric group is relabeling|check using this fact]]) and $N$ contains one of them ([[Cauchy's Theorem]]) but $4 \not{|} 6$, so $[A_{4} : N] \neq 2$. Hence $[A_{4} : N ]=3$ and so $|N|=4=2^{2}$. Then $N$ has [[the Sylow theorems|exactly one]] $2$-[[p-Sylow subgroup|sylow subgroup]] of order $4$. So, $N=K$. ![[CleanShot 2023-10-25 at 18.07.42.jpg|150]] Now, the task is to find the [[composition series]] for $K$, the [[Klein 4-group]]. $K$ has $3$ proper, nontrivial subgroups which are all [[group isomorphism|isomorphic]] to $C_{2}$; the [[quotient group|quotient]] of $K$ by each of these has order $2$ and is hence [[simple group|simple]]. So we're done: ![[CleanShot 2023-10-25 at 18.10.07.jpg|300]] > [!basicexample] Example. (Composition Series of $S_{4}$) > > $S_{4}$. First we claim that $A_{4}$ is the unique [[subgroup]] of order 12 of $S_{4}$. If $N \leq S_{4}$ has order 12 then it has index 2 (so, normal) and $S_{4} / N \cong C_{2}$ is [[abelian group|abelian]]. Thus $[S_{4}, S_{4}] \leq N$. By [[commutator series of the symmetric group]] we know that $[S_{4}, S_{4}]=A_{4}$. Because $|A_{4}|=|N|$ and $A_{4} \leq N$ we conclude that $N=A_{4}$. > Clearly $S_{4} / A_{4}$ is [[simple group|simple]], so $S_{4} \trianglerighteq A_{4}$ begins our [[composition series]]. We found the composition series of $A_{4}$ in class. Conclude that the composition series' of $S_{4}$ are the length-5 paths traversing the diagram below. ![[CleanShot 2023-10-27 at [email protected]|300]] The corresponding simple factors are $C_{2}$, $C_{3}$, and $C_{2}$. ^8a82c3 ---- #### [^1]: The reasoning, here, is that if $N$ were to contain a *single* nontrivial element of $K$, it would need to contain *all* elements of $K$ in order to remain closed under [[conjugate|conjugation]]. And, in fact, $N$ must contain an element of $K$ — else it would contain exactly $5$ 3-cycles (count) and we can show it would no longer be closed under [[conjugate|conjugation]]. ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```