----- > [!proposition] Proposition. ([[commutator series of order-12 groups]]) > [[classification of groups of order 12|Recall.]] > The [[commutator series]] for the [[group|groups]] of [[order of a group|order]] $12$ are as follows. > [!proof]- Proof. ([[commutator series of order-12 groups]]) > ># $C_{12}$ — $C_{12} \geq (e)$ Since $C_{12}$ is [[cyclic group|cyclic]], it is a [[abelian group|abelian]]. Thus $[x,y]=(e)$ for all $x,y \in C_{12}$ and we conclude that the [[commutator subgroup]] of $C_{12}$ is $[C_{12}, C_{12}]=(e)$. Thus the [[commutator series]] is $C_{12} \geq (e)$. > Since $C_{12}$ is [[abelian group|abelian]], it is [[solvable group|solvable]] by [[abelian groups are solvable]]. > ># $A_{4}$ — $A_{4} \geq C_{2} \times C_{2} \geq (e)$ The $12$ elements of $A_{4}$ are >- The $3$-cycles of $S_{4}$: $(123), (124), (132), (134), (142), (143), (234), (243)$; >- The disjoint products of $2$-cycles in $S_{4}$: $(12)(34), (13)(24), (14)(23)$; >- $e$. > >We concluded while [[composition series#^c1fedb|finding all composition series of]] $A_{4}$ that the [[Klein 4-group|Klein 4-subgroup]] of $A_{4}$, $K:=\{ e, (12)(34), (13)(24), (14)(23) \}$, is a [[normal subgroup]] of $A_{4}$. > >**Fact 1.** We also saw that all disjoint products of $2$-cycles are [[conjugate]] in $A_{4}$. > >From the [[universal property of commutator subgroup]] we know that since $A_{4} / K \cong C_{3}$ is [[abelian group|abelian]], $[A_{4}, A_{4}] \leq K$. > Since $[A_{4}, A_{4}] \neq (e)$ ($A_{4}$ not [[abelian group|abelian]]), it must in particular be the case that $[A_{4}, A_{4}]$ contains a disjoint product of $2$-cycles. By **fact 1**, it must in fact contain *all* disjoint products of $2$-cycles (since it is [[normal subgroup|normal]] in $A_{4}$). So $[A_{4}, A_{4}] \geq K$. > Conclusion: $[A_{4}, A_{4}]=K$. > Next, we have to find $[K,K]$. $K$ is [[abelian group|abelian]] so this is just $(e)$. Thus we write $A_{4} \geq C_{2} \times C_{2} \geq (e).$ Note that this is not a [[normal series]], for $C_{2} \times C_{2}$ is not [[simple group|simple]] (e.g., because it has three [[subgroup]]s of [[subgroups of index 2 are normal|index 2]]). Hence we cannot conclude from its [[commutator series]] that $A_{4}$ is [[solvable group|solvable]]. However, a [[normal series]] of $A_{4}$ is $A_{4} \trianglerighteq K \trianglerighteq (e)$ and this witnesses that it is indeed [[solvable group|solvable]]. > ># $C_{6} \times C_{2}$ — $C_{6} \times C_{2} \geq (e)$ This [[group]] is [[abelian group|abelian]] as a product of [[cyclic group]]s. So its [[commutator subgroup]] is $(e)$. It is also [[solvable group|solvable]] by [[abelian groups are solvable]]. > ># $D_{6}$ > [[dihedral group|Recall that]] $D_{6}$ is given by the relations $x^{6}=e, y^{2}=e, \text{ and } yxy^{-1}=x ^{-1},$$D_{6} = \{ e, x, x^{2},x^{3},x^{4},x^{5}, y, yx, yx^{2}, yx^{3}, yx^{4}, yx^{5} \},$ and that [[center of a group|the center]] of $D_{6}$ is $\{ e, x^{3} \}$. Also recall that $x^{k}y^{\ell}=y^{\ell}x^{-k}$. > Consider $[D_{6}, D_{6}]$. Any interesting [[commutator]] of two elements has either the form $[x^{\ell }, x^{k}]=e$ or the form $\begin{align} [x^{\ell}, yx^{k}]= & x^{\ell }y x^{k}x^{6-\ell} x^{6-k}y \\ = & x^{2\ell - 6} \\ \in & \{ e, x^{2}, x^{4} \} \end{align}$ or the form $\begin{align} [yx^{k}, x^{\ell}]= & yx^{k}x^{\ell}x^{6-k}yx^{6-\ell} \\ = & x^{-2 \ell} \\ \in & \{ e, x^{2}, x^{4} \} \end{align}$ or the form $\begin{align} [yx^{\ell}, yx^{k}]= & yx^{\ell}yx^{k}x^{6-\ell}y x^{6-k}y \\ = & x^{2k - 2 \ell} \\ \in & \{ e, x ^{2}, x^{4} \} \end{align}$ from which we conclude that every element of $[D_{6}, D_{6}]$ is a word in $e, x^{2}, \and x^{4}$. Of course, $\{ e, x^{2}, x^{4} \}=\langle x^{2} \rangle$ is a [[subgroup]] of $D_{6}$, closed under multiplication, so we conclude that $[D_{6}, D_{6}]=\langle x^{2} \rangle \cong C_{3}$. > Next we look for the [[commutator subgroup]] of $C_{3}$. But it is [[abelian group|abelian]], so its [[commutator subgroup]] is $(e)$. > Thus the [[commutator series]] for $D_{6}$ is $D_{6} \geq C_{3} \geq (e)$. > $D_{6}$ is [[solvable group|solvable]] as witnessed by the [[normal series]] $D_{4} \trianglerighteq C_{3} \trianglerighteq (e)$. > ># $C_{3} \rtimes C_{4}=\langle x, y : x^{4}=e, y^{3}=e, xyx ^{-1} = y ^{-1} \rangle$ > A general relation stemming from this presentation is $xy^{k}=y^{-k}x$. Using it we can look for [[commutator series|commutators]] in the same manner as for $D_{4}$. Any interesting [[commutator]] of two elements has either the form $[x^{\ell }, x^{k}]=e$ or ![[CleanShot 2023-10-27 at 17.45.30.jpg]] We see that the [[commutator subgroup]] is [[group isomorphism|isomorphic]] $C_{3}$. So the derived series will be $C_{3} \rtimes C_{4} \geq C_{3} \geq (e)$. It terminates at $(e)$, so the group is solvable. ^d16764 ----- #### ---- #### 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 > ```