Math 360, Fall 2021, Assignment 9
The moving power of mathematical invention is not reasoning but the imagination.
- - Augustus de Morgan
Read:[edit]
- Section 8.
Carefully define the following terms, then give one example and one non-example of each:[edit]
- Permutation (of a set $S$).
- $\mathrm{Sym}(S)$ (the symmetric group on $S$).
- $S_n$ (the symmetric group on $n$ letters).
- Order (of a group; see the text for this definition).
- $D_n$ (the $n$th dihedral group; see the text).
Carefully state the following theorems (you do not need to prove them):[edit]
- Containment criterion for subgroups of $\mathbb{Z}_n$.
- Theorem relating $\left\langle[a]\right\rangle$ to $\left\langle[\mathrm{gcd}(a,n)]\right\rangle$ (in $\mathbb{Z}_n$).
- Classification of subgroups of $\mathbb{Z}_n$ ("Every subgroup of $\mathbb{Z}_n$ is generated by a unique...").
- Theorem concerning the order of $S_n$.
Solve the following problems:[edit]
- Section 6, problems 23, 25, and 27.
- Section 8, problems 1, 3, 5, 7, 9, 11, 13, 17, 30, 31, and 33.
- In class, we listed all permutations of the set $\{1,2,3\}$. Now, make a list of all permutations of the set $\{a,b,c\}$. Do you see any relationship between the two lists?
- Suppose $S$ and $T$ are sets, and that $f:S\rightarrow T$ is a bijection. Define a function $\phi:\mathrm{Sym}(S)\rightarrow\mathrm{Sym}(T)$ by the formula $\phi(\pi)=f\circ\pi\circ f^{-1}$. Show that $\phi(\pi\circ\sigma)=\phi(\pi)\circ\phi(\sigma)$.
- With notation as above, take $S=\{1,2,3\}$ and $T=\{a,b,c\}$, and let $f:S\rightarrow T$ be given by $f(1)=a, f(2)=b, f(3)=c$. For any specific $\pi\in\mathrm{Sym}(S)$ (e.g. for $\pi=\begin{pmatrix}1&2&3\\3&2&1\end{pmatrix}$), compute the permutation $\phi(\pi)\in\mathrm{Sym}(T)$. Do this in several more specific cases.
- Returning to the general case, define a new function $\psi:\mathrm{Sym}(T)\rightarrow\mathrm{Sym}(S)$ by the formula $\psi(\tau)=f^{-1}\circ\tau\circ f$. Show that $\phi\circ\psi$ and $\psi\circ\phi$ are both the identity maps, so that $\psi$ is the inverse of $\phi$. (Note in particular that this shows that $\phi$ is an invertible function and hence is a bijection.)
- Prove that $\phi$ is an isomorphism from $(\mathrm{Sym}(S),\circ)$ to $(\mathrm{Sym}(T),\circ)$.
- Finally, show that whenever $S$ and $T$ are equinumerous sets, $\mathrm{Sym}(S)$ and $\mathrm{Sym}(T)$ are isomorphic groups.
Questions:[edit]
Solutions:[edit]
Definitions:[edit]
- Permutation (of a set $S$): Let $S$ be a set. A permutation of $S$ is a bijection from $S$ to itself.
- $\mathrm{Sym}(S)$ (the symmetric group on $S$): $U(Fun(S, S), \circ) = (Sim(S), \circ)$. That is, $Sym(S)$ means the permutation of $S$, regarded as a group under function composition. It's called the symmetric group on $S$.
- $S_n$ (the symmetric group on $n$ letters): $Sym(\{1,2,3 \cdots ,n\}) = S_n$.
- Order (of a group; see the text for this definition): the number of elements present in that group $|S|$.
- $D_n$ (the $n$th dihedral group; see the text): the group of symmetries of the regular n-gon.
Theorem:[edit]
- Containment criterion for subgroups of $\mathbb{Z}_n$: In $\mathbb Z_n, \left\langle [a] \right\rangle \subseteq \left\langle [b] \right\rangle \iff gcd(b,n) | a.$
- Theorem relating $\left\langle[a]\right\rangle$ to $\left\langle[\mathrm{gcd}(a,n)]\right\rangle$ (in $\mathbb{Z}_n$): In $\mathbb Z_n, \left\langle [a] \right\rangle = \left\langle [gcd(a,n)] \right\rangle$.
- Classification of subgroups of $\mathbb{Z}_n$ ("Every subgroup of $\mathbb{Z}_n$ is generated by a unique..."): Each subgroup has a unique generator (between $1$ and $n$) which divides $n$.
- Theorem concerning the order of $S_n$: $|S_n| = n!$.
Book Problems:[edit]
23. $<0>, <1>, <2>, <3>, <4>, <6>, <9>, <12>$ and $<18>$. (The subgroup diagram is laid out similarly to Figure 6.18, with one subgroup above another precisely when its generator divides the other's generator.)
25. $|<1>|=6/1=6, |<2>|=6/2=3, |<3>|=6/3=2, |<0>|=|<6>|=6/6=1$.
27. $|<1>|=12/1=12, |<2>|=12/2=6, |<3>|=12/3=4, |<4>|=12/4=3, |<6>|=12/6=2, |<0>|=|<12>|=12/12=1.$
1. $\begin{pmatrix}1&2&3&4&5&6\\1&2&3&6&5&4\end{pmatrix}$
3. $\begin{pmatrix}1&2&3&4&5&6\\3&4&1&6&2&5\end{pmatrix}$
5. $\begin{pmatrix}1&2&3&4&5&6\\2&6&1&5&4&3\end{pmatrix}$
7. 2
9. $\mu^2=\iota$, so $\mu^{100}=(\mu^2)^{50}=\iota^{50}=\iota.$
In problems 11 and 13, the orbit of a point is the set of all points to which it can be moved by repeated application of the permutation.
11. {1,2,3,4,5,6}.
13. {1,5}.
17. There are four choices for where to send $1$, only one choice for where to send $2$, three choices for where to send $3$, two choices for where to send $4$, and one choice for where to send $5$, so $4*1*3*2 = 24.$
30. Yes, this is a permutation. For injectivity note that $f(x_1)=f(x_2)\implies x_1+1=x_2+1\implies x_1=x_2$, and for surjectivity note that for any $y\in\mathbb{R}, f(y-1)=y$.
31. No, this map is not injective: $f(-1)=f(1)$ but $1\neq-1$.
33. Although this map is injective, it is not a permutation because it is not surjective: its image consists of the positive reals, not all reals.
Other Problems:[edit]
- $\begin{pmatrix}a&b&c\\a&b&c\end{pmatrix}, \begin{pmatrix}a&b&c\\a&c&b\end{pmatrix}, \begin{pmatrix}a&b&c\\b&a&c\end{pmatrix}, \begin{pmatrix}a&b&c\\b&c&a\end{pmatrix}, \begin{pmatrix}a&b&c\\c&b&a\end{pmatrix}, \begin{pmatrix}a&b&c\\c&a&b\end{pmatrix}$. It seems as though writing down all of the permutations of $\{1,2,3\}$ and then systematically making the replacements $1\mapsto a, 2\mapsto b, 3\mapsto c$ results in a correct list of permutations of $\{a,b,c\}$. Perhaps this correspondence might even be a group isomorphism.
- $\phi(\pi\circ\sigma)=f\circ\pi\circ\sigma\circ f^{-1}$ while $\phi(\pi)\circ\phi(\sigma)=f\circ\pi\circ f^{-1}\circ f\circ\sigma\circ f^{-1}$.
- $\begin{pmatrix}1&2&3\\a&b&c\end{pmatrix} \circ \begin{pmatrix}1&2&3\\3&2&1\end{pmatrix} = \begin{pmatrix}1&2&3\\c&b&a\end{pmatrix}$, $\begin{pmatrix}1&2&3\\c&b&a\end{pmatrix} \circ \begin{pmatrix}1&2&3\\a&b&c\end{pmatrix}^{-1}$ $= \begin{pmatrix}1&2&3\\c&b&a\end{pmatrix} \circ \begin{pmatrix}a&b&c\\1&2&3\end{pmatrix} = \begin{pmatrix}a&b&c\\c&b&a\end{pmatrix} \in Sym(T)$. By some miraculous coincidence, simply making the replacements $i\mapsto f(i)$ in the two-row notation for $\pi$ yields correct two-row notation for $\phi(\pi)$.
- For any permutation $\pi\in\mathrm{Sym}(T)$, we have $\phi(\psi(\pi))=\phi(f^{-1}\circ\pi\circ f)=f\circ f^{-1}\circ\pi\circ f\circ f^{-1}=\pi$. Similarly, for any $\sigma\in\mathrm{Sym}(S)$ we have $\psi(\phi(\sigma))=\sigma$. This shows that $\psi=\phi^{-1}$ and, in particular, that $\phi$ is a bijection.
- We showed above that $\phi$ is bijective and preserves operations, so it is an isomorphism.
- If $S$ and $T$ are equinumerous sets, then we can choose a bijection $f:S\rightarrow T$ and use it as above to produce an isomorphism $\phi:\mathrm{Sym}(S)\rightarrow\mathrm{Sym}(T)$.