Commit a243e3e7 by rrueger

### Algebra II: Lecture 8. Complete example calculations for normal subgroups of S_5

parent ab4f2b23
 ... ... @@ -259,7 +259,7 @@ groups - reach a conclusion fairly quickly. \section{Properties of Solvable Groups} \begin{theorem} \begin{theorem}\label{thm:group_solv_subgroup_solv} If a finite group $G$ is solvable, then $H \leq G$ is also solvable. \end{theorem} ... ... @@ -321,21 +321,23 @@ we must get some group theory out of the way first \begin{proof} Let $K$ be normal in $G$. We note that conjugacy defines an equivalence relation and thus induces a partition on $G$. So we have that the union of all conjugacy classes that cut $K$ non-trivially surely cover $K$ conjugacy classes that intersect $K$ non-trivially surely cover $K$ \begin{align*} K \subseteq \bigcup _{k \in K} [k] \subseteq \bigcup _{x \in G} [x] = G. \end{align*} To show the converse inclusion, $K \supseteq \cup _{k \in K} [k]$, we show for each $k \in K$ that $[k] \subseteq K$. Suppose by contradiction, that there exists some $x \in [k], x \notin K$. Then there exists some $g \in G$ such that $g k g^{-1} = x \notin K$. A contradiction to $K$ being normal. On the other hand, let $K$ be the union of conjugacy classes. By assumption $K$ is a group, we only need to check that it is normal. To that end, let $g \in G, k \in K$, then $g k g^{-1} = l$ so $k \sim l$ and thus $l \in [k] \subseteq K$. that $g k g^{-1} = x \notin K$. A contradiction to $K$ being normal. On the other hand, let $K$ be the union of conjugacy classes. By assumption $K$ is a group, we only need to check that it is normal. To that end, let $g \in G, k \in K$, then $g k g^{-1} = l$ so $k \sim l$ and thus $l \in [k] \subseteq K$. \end{proof} \begin{siderules} \begin{definition}[Cycle Structure, Reminder Algebra I] \begin{definition*}[Cycle Structure, Reminder Algebra I] The \textbf{cycle structure} of a permutation $\omega \in S_n$ is the information stored in the vector \begin{align*} ... ... @@ -343,43 +345,151 @@ we must get some group theory out of the way first \end{align*} for which $\gamma_k$ is the number of cycles of length $k$ in the cycle decomposition of $\omega$. \end{definition} \end{definition*} \begin{theorem*}[Reminder Algebra I] \begin{theorem*}[Reminder Algebra I]\label{thm:cycle_struct} Two permutations in $S_n$ are conjugates if and only if their cycle structure is the same. \end{theorem*} \end{siderules} % Now thinking back to $S_5$, we know that any proper normal subgroups $K$ of % $S_5$ must be the union of some conjugacy classes. This means that $K$ must be % the union of some of the following conjugacy classes of the representatives % \begin{align*} % \mathrm{id} & \quad |\langle (\mathrm{id}) \rangle | = 1 \\ % (1 2) & \quad |\langle (1 2) \rangle | = |S_5| = 120 \\ % (1 2 3) & \quad |\langle (1 2 3) \rangle | = {5 \choose 3} = 10 \\ % (1 2) (3 4) & \quad |\langle (1 2) (3 4) \rangle | = % \frac{1}{2} {5 \choose 2} {3 \choose 2} = 15 \\ % (1 2 3) (4 5) & \quad |\langle (1 2 3) (4 5) \rangle | = % \frac{1}{2} {5 \choose 3} = 5 \\ % (1 2 3 4) & \quad |\langle (1 2 3 4) \rangle | = {5 \choose 4} \\ % (1 2 3 4 5) & \quad |\langle (1 2 3 4 5) \rangle | = \\ % \end{align*} % We can now manually go through these, keeping in mind that if $c \in K$, then % $\langle c \rangle \subseteq K$. % \begin{itemize} % \item $K$ must contain $[\mathrm{id}]$ to be a group. % \item $K$ \textit{cannot} contain $[(1 2)]$, as otherwise $S_n = \langle (1 2) % \rangle \subseteq K$ (i.e. K will no longer be proper). % \item $K$ \textit{cannot} contain $[(1 2 3)(4 5)]$ as then $((1 2 3)(4 5))^3 = % (4 5) \in K$. Again, the span of this would be all of $S_n$. % \item $K$ \textit{cannot} contain $[(1 2 3 4)]$ as $|\langle (1 2 3 4) % \rangle| = |30|$. Together with the identity $|K| \geq 31$ % \end{itemize} Now thinking back to $S_5$, we know that any proper normal subgroups $K$ of $S_5$ must be the union of some conjugacy classes. Using Theorem~\ref{thm:cycle_struct}, we can write down the exhaustive list of all representatives of conjugacy classes in $S_5$, by simply considering all possible cycle structures % \begin{align*} [(\id)], [(1 2)], [(1 2 3)], [(1 2) (3 4)], [(1 2 3) (4 5)], [(1 2 3 4)], [(1 2 3 4 5)] \end{align*} % Now we introduce a small combinatorial result to aid our calculations % \begin{siderules} \begin{remark} Suppose we have an element $\sigma \in S_n$ with cycle structure % \begin{align*} (\gamma _1, \ldots, \gamma _n) \end{align*} % that is, it has $\gamma _k$ $k$-cycles. Then the size of the centraliser $Z(\sigma)$ is given by % \begin{align*} |Z(\sigma)| = \prod _{k=1} ^n k^{\gamma _k} \gamma _k ! \end{align*} % meaning that the size of the conjugacy class $|[\sigma]|$ is % \begin{align*} |[\sigma]| = \frac{|S_n|}{|Z(\sigma)|} = \frac{n!}{|Z(\sigma)|} \end{align*} \end{remark} % \begin{remark} This last equivalence comes about by applying the orbit-stabiliser theorem to the group action of conjugating by the group on itself. \end{remark} \end{siderules} % Using the formula above we, get that the sizes of each of these conjugacy classes are % \begin{align*} |[(\id)]| &= 1 \\ |[(1 2)]| &= 10 \\ |[(1 2 3)]| &= 20 \\ |[(1 2) (3 4)]| &= 15 \\ |[(1 2 3) (4 5)]| &= 20 \\ |[(1 2 3 4)]| &= 30 \\ |[(1 2 3 4 5)]| &= 24 \end{align*} % These clearly add up to 120 elements. So let $K$ be the normal subgroup that we are looking for, and as we know that if $K$ is a normal subgroup, it must be the union of conjugacy classes we can simply pick and choose which conjugacy classes we want inside $K$ to see if they are possible candidates. We can now set off on our search, but before we begin, we should note note one more important fact to help us sift through the possibilities. Namely, that % \begin{align*} \langle [12] \rangle = S_5 \end{align*} % that is, the transpositions generate $S_5$. Keeping in mind that if $c \in K$, then $\langle c \rangle \subseteq K$, and as we are searching for proper normal subgroups, any subgroup containing $[12]$, will generate the whole group and no longer be proper. So we can immediately say % \begin{itemize} \item $K$ \textit{cannot} contain $[(1 2)]$, as otherwise $S_5 = \langle [(1 2)] \rangle \subseteq K$ (i.e. K will no longer be proper). \item $K$ \textit{cannot} contain $[(1 2 3)(4 5)]$ as then $((1 2 3)(4 5))^3 = (4 5) \in K$. This can be done to create any transposition, meaning that all transpositions would be inside the group. Again, the span of this would be all of $S_5$. \end{itemize} % We can also make some more observations that are simply numeric, % \begin{itemize} \item $K$ \textit{cannot} contain $[(1 2 3 4)]$ as $|(1 2 3 4) | = |30|$. Together with the identity $|K| = 31$. This means we would need one more conjugacy class, as $|K|$ must divide $|S_5|$. The number of elements we can possibly add are: $29 = 60 - 31$, or $9 = 40 - 31$ to get possible divisors of $120$. Our options of conjugacy classes to include are % \begin{itemize} \item $[(123)]$. Which has 20 elements. \item $[(12)(34)]$. Which has 15 elements. \item $[(12345)]$. Which has 24 elements. \end{itemize} % on inspection, we see that none of these combinations can add to $29$ or $9$ elements. \end{itemize} % Meaning we are only left with $[(123)], [(12)(34)], [(12345)]$. These each with the identity cannot be subgroups, as they would have orders $21, 16, 25$ respectively. In combination, we see that only taking the union over all three conjugacy classes gives us a divisor of $120$, namely 60, and only closer examination, we note that this union is exactly $A_5$. We can now go on to find normal subgroups of $A_5$ using the same procedure. We show the sizes of the conjugacy classes to be % \begin{align*} |[(\id)]| &= 1 \\ |[(1 2 3)]| &= 20 \\ |[(1 2) (3 4)]| &= 15 \\ |[(1 2 3 4 5)]| &= 12 \\ |[(2 1 3 4 5)]| &= 12 \end{align*} % \begin{remark} The $2$-cycle which makes $(1 2 3 4 5)$ conjugate with $(2 1 3 4 5)$ in $S_5$ doesn't exist in $A_5$. \end{remark} % We see that adding any subset of these groups doesn't give a divisor of $|A_5| = 60$, so $A_5$ is simple and thus $S_5$ is not solvable. \begin{remark} This proof could have been significantly sped up by utilising Theorem~\ref{thm:group_solv_subgroup_solv} and noting that $A_5$ is not solvable first. \end{remark} \section{Solvability of Equations} ... ...
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