Assignment 5
- Page ID
- 169953
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- If \(\sigma \in A_n\) and \(\tau \in S_n\text{,}\) show that \(\tau^{-1} \sigma \tau \in A_n\text{.}\)
- What are the possible cycle structures of elements of \(A_5\text{?}\) What about \(A_6\text{?}\)
- Answer
-
1. Let \(\sigma \in A_n\) and \(\tau \in S_n\text{.}\)
If \(\tau \in A_n\) then \(\tau^{-1} \sigma \tau \in A_n\text{.}\) Otherwise \(\tau\) is odd permutation and \(\tau^{-1}\) is also odd (show). Then sum of odd numbers is even, \(\tau^{-1} \sigma \tau \in A_n\text{.}\)
2, The possible cycle types in \({A_5}\) are:
\(11111 , 221, 311\)
The possible cycle types in \({A_6}\) are:
\( 111111 , 2211, 3111, 33, 42 \)
- Find all of the subgroups in \(A_4\text{.}\) What is the order of each subgroup?
- What are the conjugacy classes of \(A_4\)?
- Let \(H=\{ e, (12)(34), (13)(24), (14)(23)\}.\) Find the left cosets and right cosets of \(H\) in \(A_4.\)
- Answer
-
1. Since \(|A_4|=12\), by Lagrange's theorem, the possible order of subgroups are \(1, 2, 3, 4, 6\) and \(12\). But \(A_4\) has no subgroup of order \(6\) (done in class).
Hence, the subgroups of \(A_4\) are:
Order \(1: \{e\}\)
Order \(2: \{e, (12)(34)\}, \{e, (13)(24)\}, \{e, (14)(23)\}\)
Order \(3: \{e, (123), (321)\}, \{e, (124), (421)\}, \{e, (134), (431)\}, \{e, (234), (431)\}\)
Order \(4: \{e, (12)(34), (13)(24), (14)(23)\}\)
Order \(12: A_4\)2. Show that the conjugacy classes of \(A_4\) are:
\(\{e\}, \{(12)(34), (13)(24), (14)(23)\}, \{(123), (134), (243), (142)\}, \{(321), (431), (234), (124)\}.\)3. Since \([A_4:H]=\dfrac{12}{4}=3\). Hence there are \(3\) left cosets of \(H\) in \(A_4\) and \(3\) right cosets of \(H\) in \(A_4\)
Left cosets:
\(eH = H = (12)(34)H = (13)(24)H = (14)(23)H\)
Since \((123)(12)(34) = (134)\), \((123)(13)(24) = (243)\), \((123)(14)(23) = (142),\)
\((123)H = (134)H = (243)H = (142)H = \{(123), (134), (243), (142)\}\)
Since \((132)(12)(34) = (234)\), \((132)(13)(24) = (124)\), \((132)(14)(23) = (143).\)
\((132)H = (234)H = (124)H = (143)H = \{(132), (234), (124), (143)\}\)
Right cosets:
\(He = H = H(12)(34) = H(13)(24) = (14)(23)H\)
Since \((12)(34)(123) = (243)\), \((13)(24)(123) = (142)\), \((14)(23)(123) = (134),\)
\(H(123) = H(243) = H(142) = H(134) = \{(123), (243), (142), (134)\}\)
Since \((12)(34)(132) = (143)\), \((13)(24)(132) = (234)\), \((14)(23)(132) = (124),\)
\((132)H = (143)H = (234)H = (124)H = \{(132), (143), (234), (124)\}.\)
- Find the center of \(D_4\) and \(D_5\). Justify your answer.
- What is the center of \(D_n\text{?}\)
- Find the conjugacy classes of \(D_4\). Justify your answer.
- Answer
-
1. Consider \( a = s^r k \) and \( b = r^j \) in \( D_n \) for some \( 0 \leq k, j < n \). Then
\[
ab = s^{r(k+j)},
\]
and,
\[
ba = s^{r(k-j)}.
\]
The equation \( ab = ba \) implies that
\[
k - j \equiv k + j \pmod{n}.
\]
But this is only true if \( j = \frac{n}{2} \) or \( j = 0 \).
Therefore, all reflections are not in the center of \( D_n \). Furthermore, the only rotation that is in the center is \( r^{\frac{n}{2}} \) (and trivially \( e \)).
So, if \( n \) is odd, then the center of \( D_n \) is \( \{ e \} \), otherwise, the center of \( D_n \) is \( \{ e, r^{\frac{n}{2}} \} \).
For example,
\[
Z(D_4) = \{ e, r^2 \}, \quad Z(D_5) = \{ e \}.
\]2. The center of \(D_n\) is \(\{e\}\) if \(n\) is odd and \(\{e, r^{n/2}\}\) if \(n\) is even.
3.
Let \(D_4 = <r, s | r^4 = e, s^2 = e, srs = r^{-1}>.\)
The conjugacy classes of \(D_4\) are \(\{e\}, \{r^2\}, \{r,r^3\}, \{s, r^2s\}, \{rs, r^3s\}\).
Clearly, \([e] = \{e\}\).Consider the following for the class \([r^m]\) for some \(m\):
\[
r^n r^m r^{-n} = r^m,
\]
and
\[
s r^n r^m r^{-n} s = r^{-m},
\]
for all \(n\). Therefore, \([r^m] = \{r^m, r^{-m}\}\). That is,
\[
[r] = \{r, r^3\}, \quad [r^2] = \{r^2\}.
\]Consider the class \([sr^m]\) for some \(m\). We have:
\[
r^n s r^m r^{-n} = s r^{m - 2n},
\]
and
\[
s r^n s r^m s r^{-n} = s s r^{-n + m + n} = s r^{-m}.
\]
for all \(n\). Since \(D_4\) has an even number of rotations, we can partition the reflections into two conjugacy classes: even and odd. If there were only one conjugacy class of reflections, \(m - 2n\) would not cycle properly. Thus, we have:
\[
[s r^2] = \{ s r^n \mid n \in 2\mathbb{N} \} = \{s, sr^2\},
\]
and
\[
[s r] = \{ s r, s r^3 \}.
\]To summarize, the conjugacy classes are:
\[
[e], [r], [r^2], [s], [sr].
\]
Thus, the members of a conjugacy class of \( D_4 \) are different but have the same type of effect on a square:- \( r \) and \( r^3 \) are a 90-degree rotation in some direction,
- \( s \) and \( r^2s \) are a reflection across a diagonal,
- \( rs \) and \( r^3s \) are a reflection across an edge bisector.
List all of the subgroups of \(S_4\text{.}\) Find each of the following sets:
- \(\displaystyle \{ \sigma \in S_4 : \sigma(1) = 3 \}\)
- \(\displaystyle \{ \sigma \in S_4 : \sigma(2) = 2 \}\)
- \(\{ \sigma \in S_4 : \sigma(1) = 3\) and \(\sigma(2) = 2 \}\text{.}\)
- Are any of these sets subgroups of \(S_4\text{?}\)
Let \(\sigma \in S_X\text{.}\) If \(\sigma^n(x) = y\) for some \(n \in \mathbb Z\text{,}\) we will say that \(x \sim y\text{.}\)
- Show that \(\sim\) is an equivalence relation on \(X\text{.}\)
- Define the orbit of \(x \in X\) under \(\sigma \in S_X\) to be the set
\[ {\mathcal O}_{x, \sigma} = \{ y : x \sim y \}\text{.} \nonumber \]Compute the orbits of each element in \(\{1, 2, 3, 4, 5\}\) under each of the following elements in \(S_5\text{:}\)
\begin{align*} \alpha & = (1254)\\ \beta & = (123)(45)\\ \gamma & = (13)(25)\text{.} \end{align*}
That is for example: \begin{align*} {\mathcal O}_{x, \sigma}& = \{ y \in \{1,2,3,4,5\} : x \sim y \}\\
&=\{ y :\sigma^n(x) = y,\text{for some } n\in \mathbb{N} \}\\
\end{align*} -
If \({\mathcal O}_{x, \sigma} \cap {\mathcal O}_{y, \sigma} \neq \emptyset\text{,}\) prove that \({\mathcal O}_{x, \sigma} = {\mathcal O}_{y, \sigma}\text{.}\) The orbits under a permutation \(\sigma\) are the equivalence classes corresponding to the equivalence relation \(\sim\text{.}\)
A subgroup \(H\) of \(S_X\) is transitive if for every \(x, y \in X\text{,}\) there exists a \(\sigma \in H\) such that \(\sigma(x) = y\text{.}\) Prove that \(\langle \sigma \rangle\) is transitive if and only if \({\mathcal O}_{x, \sigma} = X\) for some \(x \in X\text{.}\)