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Groups and group actions problem sheet 6

 
  • Feb 01, 2022
  1. (i) Explain why there is only one homomorphism from $\mathbb{Z}_{5}$ to $D_{8}$ but five homomorphisms from $\mathbb{Z}_{5}$ to $D_{10}$.
    (ii) What are the normal subgroups $N$ of $S_{3}$ ? For each $N$, identify the quotient group $S_3 / N$.
    (iii) Hence, or otherwise, explain why there are two homomorphisms from $S_3$ to $\mathbb{Z}_6$ and six homomorphisms from $S_3$ to $D_8$.
    Solution.
    (i)Let $\phi$ be a homomorphism from $\Bbb{Z}_5$ to $D_8$. By first isomorphism theorem, $\Bbb{Z}_5/\kerϕ≌\operatorname{im}\phi$, so $5=\lvert\ker \phi\rvert\lvert\operatorname{im}\phi\rvert$. Since $|\operatorname{im}\phi|$ divides $5$ and $8$, $|\operatorname{im}\phi|=1$. Hence $\phi$ is the trivial map which sends every elements of $\Bbb{Z}_5$ to identity of $D_8$.  
    For the second part, write $\Bbb{Z}_5=\langle x\rangle$ and $D_{10}=\langle r,s|r^5=s^2=(rs)^2=1\rangle$.  
    $o(ϕ(x))|o(x)=5$, so $\operatorname{im}ϕ=⟨r⟩,ϕ(x)=r^i,i=0,1,2,3,4$, so we have five homomorphisms.
    (ii)Conjugacy classes of $S_3$ have size 1,2,3 and a subgroup should contain $e$, we have $1|6,1+2|6,1+3\nmid6,1+2+3|6$, so the normal subgroups of $S_3$ are $\{e\},A_3,S_3$, and $S_3/\{e\}≌S_3,S_3/A_3≌C_2,S_3/S_3≌\{e\}$.
    (iii)A homomorphism $ϕ$ from $S_3$ to $\Bbb Z_6$ has kernel $\{e\},A_3,$ or $S_3$.
    If $\kerϕ=\{e\}$, then $|\operatorname{im}ϕ|=6$, so $\operatorname{im}ϕ=\Bbb Z_6$ is abelian but $S_3$ is non-abelian, so no such $ϕ$.
    If $\kerϕ=A_3$, then $|\operatorname{im}ϕ|=2$, so $\operatorname{im}ϕ=\{\bar0,\bar3\}$, so $\phi$ sends even permutations to $\bar0$ and odd ones to $\bar3$.
    If $\kerϕ=\Bbb Z_6$, then $|\operatorname{im}ϕ|=1$, so $ϕ$ is the trivial homomorphism.
    For the second part, a homomorphism $ϕ$ from $S_3$ to $D_8$ has kernel $\{e\},A_3,$ or $S_3$.
    If $\kerϕ=\{e\}$, then $|\operatorname{im}ϕ|=6\nmid8$, so no such $ϕ$.
    If $\kerϕ=A_3$, then $|\operatorname{im}ϕ|=2$. Let $D_8=\{e,r,r^2,r^3,s,rs,r^2s,r^3s\}$, so $\operatorname{im}ϕ$ can be $\{e,r^2\},\{e,rs\},\{e,r^2s\},\{e,r^3s\}$, so $\phi$ sends even permutations to $\operatorname{im}ϕ$ and odd ones to $D_8∖\operatorname{im}ϕ$.
    If $\kerϕ=\Bbb Z_6$, then $|\operatorname{im}ϕ|=1$, so $ϕ$ is the trivial homomorphism.
  2. Let $G$ be a group and let $N$ be a normal subgroup of $G$. Let $H$ be a subgroup of $G$ with $H \geqslant N$.
    (i) Show that $H / N=\{h N \mid h \in H\}$ is a subgroup of $G / N$.
    (ii) Assume further that $H$ is a normal subgroup of $G$. Show that $H / N$ is a normal subgroup of $G$.
    Solution.
    (i)Clearly, $H/N⊂G/N$. For any $h_1,h_2∈H,(h_1^{-1} N)(h_2N)=(h_1^{-1}h_2)N∈H/N$. Therefore $H/N≤G/N$.
    (ii)For any $g∈G$, because $H⊲G$, we have $ghg^{-1}=h'∈H$; because $N⊲G$, we have $gNg^{-1}=N$.$$g(H/N)g^{-1}=\{ghNg^{-1}:h∈H\}=\{(ghg^{-1})(gNg^{-1}):h∈H\}=\{h'N:h∈H\}⊂H/N$$Therefore $H/N⊲G$.
  3. Show that the group $S_n$ (right) acts on the set of subsets of $\{1,2, \ldots, n\}$ by $\rho(S, \sigma)=S \sigma$.
    Show that there are $n+1$ orbits, one for each possible value of $|S|$. Show that if $|S|=k$ then $\operatorname{Stab}(S) \cong S_{k} \times S_{n-k}$.
    Solution.
    For all $S⊂\{1,2,⋯,n\}$, $ρ(S,{\rm id})=S\,{\rm id}=S$, $ρ(ρ(S,σ_1),σ_2)=(Sσ_1)σ_2=S(σ_1σ_2)=ρ(S,σ_1σ_2)$.
    $|S|=|Sσ|$. Conversely, for two subsets $\{a_i\},\{b_i\}$ of the same size $k∈\{0,1,⋯,n\}$, the permutation $σ=\pmatrix{a_1&⋯&a_k\\b_1&⋯&b_k}$ satisfies $ρ(\{a_i\},σ)=\{b_i\}$, so there are $n+1$ orbits.
    If $|S|=k$ and $σ∈S_n$ such that $Sσ=S$, then $σ=σ_1σ_2$ where $σ_1∈\operatorname{Sym}S≌S_k$ and $σ_2∈\operatorname{Sym}(\{1,⋯,n\}∖S)≌S_{n-k}$. Conversely, for $σ_1∈\operatorname{Sym}S$ and $σ_2∈\operatorname{Sym}(\{1,⋯,n\}∖S)$, we have $Sσ_1σ_2=S$. Therefore $\operatorname{Stab}(S) \cong S_{k} \times S_{n-k}$.
  4. Let $S$ denote the set of possible black-or-white colourings of the edges of an equilateral triangle. Explain why $|S|=8$.
    The triangle's symmetry group $D_6$ acts naturally on $S$. How many orbits are there?
    By listing an element from each orbit, show that there are 10 orbits if three colours are used.
    Solution.
    Because each edge is black or white, number of colorings$=2^3=8$.
    There are 4 orbits: each consists of colorings of $k$ white edges and $3-k$ black ones, $k=0,1,2,3$.
    Denote three colors A,B,C, then a list of an element from each orbit is AAA,BBB,CCC,ABB,ACC,BAA,BCC,CAA,CBB,ABC.
  5. (i) Show that $V_4=\{e,(12)(34),(13)(24),(14)(23)\}$ is a normal subgroup of $S_4$.
    (ii) Let $H=\operatorname{Sym}\{1,2,3\}=\left\{σ∈S_4:4σ=4\right\}$. How many left cosets of $H$ are there?
    Show that if $h_1,h_2∈H$ and $h_1V_4=h_2V_4$ then $h_1=h_2$.
    (iii) Deduce that $S_4/V_4$ is isomorphic to $S_3$.
    (iv) With the aid of diagrams, determine the number of essentially different ways there are of labelling the vertices of a rectangle as $1,2,3,4$. How do these diagrams relate to the cosets of $V_4$ in $S_4$?
    Solution.
    (i) First, $V_4=⟨(12)(34),(13)(24)⟩$ is a subgroup of $S_4$. $V_4$ is a union of conjugacy classes $\{e\},\{(12)(34),(13)(24),(14)(23)\}$, so $V_4$ is a normal subgroup of $S_4$.
    (ii) The number of left cosets of $H$ is $|S_4|/|H|=4!/3!=4$.
    $h_1V_4=h_2V_4⇒∃v∈V_4:h_2=h_1v$. Therefore $4=4h_2=4h_1v=4v$. But the only element $v∈V_4$ that fixes $4$ is $v = e$, so $h_2=h_1e=h_1$.
    (iii) Since $\lvert S_4/V_4\rvert=4!/4=6$, by (ii) six cosets $hV_4,h∈H$ are all distinct, so $S_4/V_4≌H≌S_3$.
    (iv) $S_4$ generates all possible labelings. $V_4$ is symmetries of the rectangle, orbits are essentially different labeling. $S_4/V_4$ identifies not essentially different labelings. $H$ is relabellings of 123. Every essentially different labelling by permuting 123 in rectangle, and every permutation gives essentially different labeling, as in the figures below.
          

    Starter
    S1. By constructing a homomorphism with the appropriate kernel and image and then applying the Isomorphism Theorem, demonstrate the following isomorphisms:
    (i) $\mathbb{R}^*/\{±1\}≌(0,∞)$.
    (ii) $\mathbb{C}^*/\{±1\}≌\mathbb{C}^*$.
    (iii) $\mathbb{C}^*/S^1≌(0,∞)$.
    (iv) $\mathbb{R}/\mathbb{Z}≌S^1$.
    Solution.
    (i) $ϕ:\mathbb{R}^{*}→(0,∞),ϕ(x)=x^2$ is a homomorphism, $\kerϕ=\{±1\}$.
    (ii) $ϕ:\mathbb{C}^{*}→\mathbb{C}^*,ϕ(x)=x^2$ is a homomorphism, $\kerϕ=\{±1\}$.
    (iii) $ϕ:\mathbb{C}^{*}→(0,∞),ϕ(x)=|x|$ is a homomorphism, $\kerϕ=S^1$.
    (iv) $ϕ:\mathbb{R}→S^1,ϕ(x)=e^{2πix}$ is a homomorphism, $\kerϕ=\mathbb Z$.

    Pudding
    P1. (i) Let $G$ be a group and $H$ a normal subgroup of $G$ with $|H|=n$. Further let $g∈G$ be such that $gH$ has order $m$ in $G/H$.
    Show that $⟨g⟩H=\left\{g^kh:k∈\mathbb{Z},h∈H\right\}$ is a subgroup of $G$ of order $mn$.
    (ii) Use this method to construct subgroups of $S_4$ of orders 8 and 12.
    Solution.
    (i) Clearly $e∈⟨g⟩H$. For $k_1,k_2∈\mathbb{Z},h_1,h_2∈H$, $(g^{k_1}h_1)^{-1}g^{k_2}h_2=h_1^{-1}g^{-k_1+k_2}h_2=g^{-k_1+k_2}(h_3h_2)∈⟨g⟩H$ where $h_3=g^{k_1-k_2}h_1^{-1}g^{-k_1+k_2}∈H$ because $H⊲G$. By subgroup criterion, $⟨g⟩H≤G$.
    What is the order of $⟨g⟩H$?
    Have that $g H$ has order $m$ in $G / H$, so $(g H)^{m}$ is the identity in $G / H$, so $g^{m} \in H$, and moreover $g^{k} \notin H$ for $1 \leqslant k \leqslant m-1$ (or $g H$ would have smaller order in $G/H$).
    Consider $\left\{g^k h: 0 \leqslant k \leqslant m-1, h \in H\right\}$. This must be $\langle g\rangle H$.
    Can two of its elements be the same?
    If $g^{k} h=g^{\ell} h^{\prime}$, WLOG $k \leqslant \ell$,
    then $h=g^{\ell-k} h^{\prime} \in H$
    so $g^{\ell-k} \in H$, so $k=\ell$.
    So $\langle g\rangle H=\left\{g^{k} h: 0 \leqslant k \leqslant m-1, h \in H\right\}$, and this contains $m n$ distinct elements.
    (ii) Consider $G=S_{4}$.
    We have $V_{4} \unlhd S_{4}$ (see Q5), and $\left|V_{4}\right|=4$.
    What is the order of $(12) V_{4}$ in $S_{4} / V_{4}$ ?
    Well, clearly at most 2 , and $(12) V_{4} \neq V_{4}$ (as $(12)((12)(34)) \notin V_{4}$), so it has order 2 in $S_{4} / V_{4}$. So, by (i), $\langle(12)\rangle V_{4}$ is a subgroup of $S_{4}$ of order 8 .
    What is the order of $(123) V_{4}$ in $S_{4} / V_{4}$ ?
    A quick check shows that it's 3 .
    So, by (i), $\langle(123)\rangle V_{4}$ is a subgroup of $S_{4}$ of order 12 .

    Supplement
    Show that not all 5 cycles are conjugate in $A_5$.
    Solution. https://math.stackexchange.com/q ... re-conjugate-in-a-5
    The direct approach is to find two non-conjugate 5-cycles in $A_5$, so we need to:
    • Start with a 5-cycle $(a\,b\,c\,d\,e)$
    • Show that there exists another 5-cycle $(f\,g\,h\,i\,j)$ such that for all $ρ∈A_5$ we have $(aρ\,bρ\,cρ\,dρ\,eρ)≠(f\,g\,h\,i\,j)$
    This is a rather cumbersome approach, though, especially since there are 5 equivalent but distinct ways to represent a given 5-cycle with this notation.
    If you know that the order of a conjugacy class must divide the order of the group, then we can proceed much more simply. How many 5-cycles are there? 4!=24 (keep 1 fixed, and permute 4 others). What is the order of $A_5$? 5!/2=60 (half of permutations are even). And since 24 is bigger than 20, the largest proper divisor of 60, then it follows that there exist non-conjugate 5-cycles—with no need to mention any specifically. Incidentally, there are exactly 2 conjugacy classes of 5-cycles in $A_5$, those conjugate to (1 2 3 4 5) and those conjugate to (2 1 3 4 5).
    Remark. By orbit-stabilizer theorem, the size of conjugacy class $[g]$ of the element $g$ in a finite group is$$|[g]|={|G|\over|C_G(g)|}\tag⋆$$where $C_G(g)$ is the centralizer of $g$ in $G$. Our knowledge about conjugacy classes in $S_5$ tells us that a 5-cycles has 24 conjugates in $S_5$. Let $\sigma\in S_5$ be a fixed 5-cycle. Formula (⋆) tells us that $C_{S_5}(\sigma)$ is of order five. As any element commutes with its powers, we can conclude that$$C_{S_5}(\sigma)=\langle \sigma\rangle.$$As this subgroup is contained in $A_5$, we can conclude that $$C_{A_5}(\sigma)=C_{S_5}(\sigma).$$Another application of Formula (⋆) then tells us that the conjugacy class of any 5-cycle in $A_5$ has exactly 12 elements. In other words, the $S_5$ conjugacy class splits in two.
    More detailed information can be obtained by studying the normalizer of a Sylow $5$-subgroup in $S_5$. There are six such Sylow subgroups, so their normalizers have order $20$. If we specify $\sigma=(12345), P_5=\langle\sigma\rangle$ we see that the 4-cycle $\beta=(2354)$ satisfies the relation$$\beta\sigma\beta^{-1}=[1\mapsto\beta(2)\mapsto\beta(3)\mapsto\beta(4)\mapsto\beta(5)\mapsto1]=(13524)=\sigma^2.$$Therefore $N_{S_5}(P_5)=\langle\sigma,\beta$ as the latter clearly normalizes $P_5$ and it has the correct order 20.

    The problem comes from the fact that $\beta\notin A_5$. Any element $\tau\in S_5$
    with the property $\tau\sigma\tau^{-1}=\sigma^2$ has the property that $\tau\beta^{-1}$ must commute with $\sigma$. In other words we must have $\tau=\beta\sigma^k$ for some $k$. But all these permutations are odd and thus not available in $A_5$.