4.4: Alternating Groups
- Page ID
- 100564
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section, we describe a special class of permutation groups. To get started, let’s play with a few exercises.
Write down every permutation in \(S_3\) as a product of 2-cycles in the most efficient way you can find (i.e., use the fewest possible transpositions). Now, write every permutation in \(S_3\) as a product of adjacent 2-cycles, but don’t worry about whether your decompositions are efficient. Any observations about the number of transpositions you used in each case? Think about even versus odd.
If \(\alpha_1,\alpha_2,\ldots,\alpha_k\) is a collection of 2-cycles in \(S_n\) such that \(\alpha_1\alpha_2\cdots\alpha_k=(1)\), then \(k\) must be even.
- Proof.
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Suppose \(\alpha_1,\alpha_2,\ldots,\alpha_k\) is a collection of 2-cycles in \(S_n\) such that \(\alpha_1\alpha_2\cdots\alpha_k=(1)\). We need to show that \(k\) is even. We proceed by strong induction. First, it is clear that the statement is not true when \(k=1\), but is true when \(k=2\).
Now, assume that \(k>2\) and if \(j\leq k-1\) and we have a product of \(j\) 2-cycles that equals the identity, then \(j\) is even. Consider \(\alpha_1\alpha_2\). The only possibilities are:
- \(\alpha_1\alpha_2=(a,b)(a,b)\),
- \(\alpha_1\alpha_2=(a,b)(a,c)\),
- \(\alpha_1\alpha_2=(a,b)(c,d)\),
- \(\alpha_1\alpha_2=(a,b)(b,c)\).
If case (i) happens, then \[(1)=\alpha_1\alpha_2\cdots\alpha_k=\alpha_3\alpha_4\cdots\alpha_k.\] Since the expression on the right consists of \(k-2\) factors, \(k-2\) must be even by induction, which implies that \(k\) is even. Now, suppose we are in one of cases (ii), (iii), or (iv). Observe that:
- \((a,b)(a,c)=(b,c)(a,b)\),
- \((a,b)(c,d)=(c,d)(a,b)\),
- \((a,b)(b,c)=(b,c)(a,c)\).
In each case, we were able to move \(a\) from the original left 2-cycle to a new right 2-cycle. That is, we were able to rewrite \(\alpha_1\alpha_2\) so that \(a\) does not appear in the left 2-cycle. Systematically repeat this process for the pairs \(\alpha_2\alpha_3\), \(\alpha_3\alpha_4\),…, \(\alpha_{k-1}\alpha_k\). If we ever encounter case (i) along the way, then we are done by induction. Otherwise, we are able to rewrite \(\alpha_1\alpha_2\cdots\alpha_k\) so that \(a\) only appears in the rightmost 2-cycle. But this implies that \(\alpha_1\alpha_2\cdots\alpha_k\) does not fix \(a\), which contradicts \(\alpha_1\alpha_2\cdots\alpha_k =(1)\). This implies that at some point we must encounter case (i), and hence \(k\) is even by induction.
If \(\sigma\in S_n\), then every transposition representation of \(\sigma\) has the same parity.
The previous theorem tells us that the following definition is well-defined.
A permutation is even (respectively, odd) if one of its transposition representations consists of an even (respectively, odd) number of transpositions.
Classify all of the permutations in \(S_3\) as even or odd.
Classify all of the permutations in \(S_4\) as even or odd.
Determine whether \((1,4,2,3,5)\) is even or odd. How about \((1,4,2,3,5)(7,9)\)?
Consider the arbitrary \(k\)-cycle \((a_1,a_2,\ldots, a_k)\) from \(S_n\) (with \(k\leq n\)). When will this cycle be odd versus even? Briefly justify your answer.
Conjecture a statement about when a permutation will be even versus odd. Briefly justify your answer.
And finally, we are ready to introduce the alternating groups.
The set of all even permutations in \(S_n\) is denoted by \(A_n\) and is called the alternating group.
Since we referred to \(A_n\) as a group, it darn well better be a group!
The set \(A_n\) forms a group under composition of permutations and has order \(n!/2\).*
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Hint: To show that \(A_n\) is a group, argue that \(A_n\) is a subgroup of \(S_n\) using the Two-Step Subgroup Test (see Theorem [thm:subgroup_criterion]). To show that \(|A_n|=n!/2\), prove that the number of even permutations in \(S_n\) is the same as the number of odd permutations in \(S_n\). Here is one way to accomplish this. Define \(f:A_n\to S_n\setminus A_n\) via \(f(\sigma)=(1,2)\sigma\). Note that \(S_n\setminus A_n\) is the set of odd permutations in \(S_n\). Show that \(f\) is a bijection.
Find \(A_3\). What group is \(A_3\) isomorphic to?
Find \(A_4\) and then draw its subgroup lattice. Is \(A_4\) abelian?
What is the order of \(A_5\)? Is \(A_5\) abelian?
What orders of elements occur in \(S_6\) and \(A_6\)? What about \(S_7\) and \(A_7\)?
Does \(A_8\) contain an element of order 15? If so, find one. If not, explain why no such element exists.
Below are a few interesting facts about \(A_4\) and \(A_5\), which we will state without proof.
- The group of rigid motion symmetries for a regular tetrahedron is isomorphic to \(A_4\).
- You can arrange the Cayley diagram for \(A_4\) with generators \((1,2)(3,4)\) and \((2,3,4)\) on a truncated tetrahedron, which is depicted in Figure \(\PageIndex{1}\)a.
- You can arrange the Cayley diagram for \(A_5\) with generators \((1,2)(3,4)\) and \((1,2,3,4,5)\) on a truncated icosahedron, which is given in Figure \(\PageIndex{1}\)b. You can also arrange the Cayley diagram for \(A_5\) with generators \((1,2,3)\) and \((1,5)(2,4)\) on a truncated dodecahedron seen in Figure \(\PageIndex{1}\)c.
