Group theory problem sheet 1

Claim: $H⩽G$ [Prop 40] Proof: By definition $H⊂G$, for a subgroup we also need: • identity: $eK=K∈\bar H⇒e∈H$. • inverses: $∀g∈H:gK∈\bar H⇒(gK)^{-1}=g^{-1}K∈\bar H⇒g^{-1}∈H$. • closure: $∀g_1,g_2∈H:g_1K, g_2K∈\bar H⇒g_1Kg_2K=g_1g_2K∈\bar H⇒g_1g_2∈H$. Claim: $K⊴H,H/K=\bar{H}$ Proof: Since $e∈H$, for all $k∈K,kK=K=eK∈\bar H⇒k∈H⇒K⊂H⇒K⩽H$. Since $K⊴G,H⩽G$, for all $h∈H,k∈K$, we have $h^{-1}kh∈K$, so $K⊴H$. By the definition of π, $H/K=π(H)=π(π^{-1}(\bar{H}))=\bar{H}$. Claim: if $\bar{H}⊴G/K$ then $H⊴G$ Proof: We proved $H⩽G$, so only need to prove closure under conjugation. Let $h∈H=π^{-1}(\bar{H})$ and $g∈G$. Then by normality of $\bar{H}$ in $G/K$ we have $$(gK)^{-1}hK(gK)=g^{-1}hgK∈\bar{H}.$$Therefore, $g^{-1}hg∈π^{-1}(\bar{H})=H$ as required. For any word $w$ in $a$ and $b$, using $ba=a^{-1}b$, we can move every $a$ after $b$, so $w=b^ia^j$ for some $i,j∈ℤ$. Using $a^n=b^2=1$, reduce to $i∈\{0,1\},j∈\{0,1,…,n-1\}$, so $G$ has ≤ 2n elements. Suppose $b^{i_1}a^{j_1}=b^{i_2}a^{j_2}$, then $b^{i_1-i_2}=a^{j_2-j_1}$, so $2|i_1-i_2,n|j_2-j_1$, so these $2n$ elements of $G$ are distinct. To prove $\left(b a^i\right)^2=1$ by induction, for $i=0$, $\left(b a^i\right)^2=b^2=1$. Suppose it is true for $i-1$, we have $$ba^iba^i=ba^i(ba)a^{i-1}=ba^i(a^{-1}b)a^{i-1}=ba^{i-1}ba^{i-1}=1$$so $\left(b a^i\right)^2=1$ for all $i∈ℕ$, using $a^n=1$ it is true for all $i∈ℤ$. $D_{2n}$ is symmetry group of regular $n$-gon: $a^i$ are rotations (of order $n$), $a^ib$ are reflections (of order 2).

1. for any $x^i\,(i∈ℤ)$, let $k=i\bmod 6\,(k∈\{0,…,5\})$, then $x^i=x^k$, and $x^k(k∈\{0,…,5\})$ are distinct, so $G_1≅C_6$
2. using $xy=yx$, every element of $G_2$ can be written as $x^iy^j\,(i,j∈ℤ)$ and they are distinct, so $G_2≅ℤ^2$
3. $x^3y=x^2y⇒x=1$, then $y=y^2⇒y=1$, so $G_3=\{1\}$
4. $G_4$ is the quotient of $G_2=ℤ^2$ by $⟨(5,-3)⟩$. Consider the homomorphism $f:ℤ^2→ℤ,(a,b)↦3a+5b$. Since $3×-3+5×2=1$, $f$ is surjective. Let $(a,b)∈\ker f$, then $0=3a+5b⇒5|3a+5b⇒5|a$. Let $a=5t$, then $b=-3t⇒\ker f=⟨(5,-3)⟩$ By first isomorphism theorem, $G_4≅ℤ$
5. $G_5$ is the quotient of $G_2=ℤ^2$ by $⟨(4,-2)⟩$. Consider the homomorphism $f:ℤ^2→ℤ⊕ℤ_2,(a,b)↦(a+2b,b\bmod2)$ For any $(a,b)∈ℤ⊕ℤ_2$, we have $(a,b)=f(a-2b,b)$, so $f$ is surjective. Let $(a,b)∈\ker f$, then $a+2b=b\bmod2=0$. Let $b=-2t$, then $a=4t⇒\ker f=⟨(4,-2)⟩$ By first isomorphism theorem, $G_5≅ℤ⊕ℤ_2$

1. An element $w$ of $G$ is a finite word in $x$ and $y$, after eliminating $xx,yy$, if $w$ ends with $y$, either $w=z^k$ or $yz^k$; if $w$ ends with $x$, $wz$ ends with $y$, so $wz=z^k$ or $wz=yz^k$, so $w=z^{k-1}$ or $w=yz^{k-1}$.
2. To verify $y↦y,z↦xy$ is a homomorphism $D_∞→G$: $y^2=1$ and $yzy^{-1}=yxyy^{-1}=yx=z^{-1}$ To verify $y↦y,x↦zy$ is a homomorphism $G→D_∞$: $y^2=1$ and $x^2=zyzy=z(yzy^{-1})=zz^{-1}=1$
3. The following is an isomorphism from $G$ to the isometry group of ℤ as a subset of ℝ\begin{array}{l|l}x&n↦1-n&\text{reflection about 1/2}\\y&n↦-n&\text{reflection about 0}\\z=xy&n↦1+n&\text{translation by 1}\end{array}

1. $G$ is nonabelian⇒ $G$ is not cyclic⇒ no element has order 8. If $G$ has no elements of order 4, by Lagrange's theorem, every element of $G$ has order 1 or 2. For all $x,y∈G$, $yx=y^{-1}x^{-1}=(xy)^{-1}=xy⇒G$ is abelian.
2. The cosets $A,Ab$ (each of size 4) partition $G$. Since $b∉A$ we have $ba∉A⇒ba∈Ab=\{b,ab,a^2b,a^{-1}b\}$. If $ba=b$, then $a=1$, contradiction; If $ba=a^2b$, then $a=b^{-1}a^2b⇒a^2=b^{-1}a^4b=1$, contradiction (4 = order of a); If $ba=ab$, since $G=⟨a,b⟩$, $G$ is abelian, contradiction. Therefore $ba=a^{-1}b⇒b a b^{-1}=a^{-1}$ Since $b∉A$, we have $b^2∉Ab⇒b^2∈A=\{1,a,a^2,a^{-1}\}$. If $b^2$ is $a$ or $a^{-1}$, then $b$ has order 8, contradicting (i). So $b^2=1$ or $a^2$.
3. If $b^2=1,a^4=1,b a b^{-1}=a^{-1}$, by Q2, $a^ib^j↦(i,j)$ is an isomorphism $G→D_8$ If $b^2=a^2,a^4=1,b a b^{-1}=a^{-1}$, can verify $a^ib^j↦(i,j)$ is an isomorphism $G→Q_8$ So there are 2 non-Abelian groups of order 8. Since $8=2^3$, there are 3 Abelian groups of order 8: $C_2^3,C_4×C_2,C_2×C_2×C_2$. [classification of Finitely generated abelian group]

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   $\operatorname{Aut}(V_4)≅S_3$ $\operatorname{Aut}(Q_8)≅V_4⋊S_3$ $S_n≅\operatorname{Aut}S_n$ except $n=6$

Theorem 3.8. For $n$ odd, the normal subgroups of $D_{2n}$ are $D_{2n}$ and $⟨a^d⟩$ for all $d∣n$. For $n$ even, $D_{2n}$ have two more normal subgroups $⟨a^2,b⟩$ and $⟨a^2,ab⟩$.

$C_{18}=C_9×C_2$ has 3 comp. series: $\{e\}◃C_3◃C_9◃C_{18}$ $\{e\}◃C_2◃C_6◃C_{18}$ $\{e\}◃C_3◃C_6◃C_{18}$ Composition factors: $C_2,C_2,C_3$.

$D_{10}=C_5⋊C_2$ has 2 comp. series: $\{e\}◃C_5◃D_{10}$ $\{e\}◃C_2◃D_{10}$ Composition factors: $C_2,C_5$.

$D_8=C_4⋊C_2$ has 2 comp. series: $\{e\}◃C_2◃C_4◃D_8$ $\{e\}◃C_2◃V_4◃D_8$ Composition factors: $C_2,C_2,C_2$. two $V_4⊂D_8$: $⟨a^2,b⟩,⟨a^2,ab⟩$

$Q_8$ has 1 comp. series: $\{e\}◃C_2◃C_4◃Q_8$ Composition factors: $C_2,C_2,C_2$. $C_4$ has 3 realizations $⟨i⟩,⟨j⟩,⟨k⟩$

$C_4×C_2$ has 2 comp. series: $\{e\}◃C_2◃C_4◃C_4×C_2$ $\{e\}◃C_2◃C_2×C_2◃C_4×C_2$ (⇒) If $HK⩽G$, for any $h∈H,k∈K$, since $h^{-1}∈H,k^{-1}∈K$, we have $h^{-1}k^{-1}∈HK$, so $kh=(h^{-1}k^{-1})^{-1}∈HK$, so $HK⊃KH$, similarly $HK⊂KH$, so $HK=KH$. (⇐) If $HK=KH$, for any $h_1,h_2∈H,k_1,k_2∈K$ we have $k_1^{-1}h_1^{-1}h_2∈KH$, so $k_1^{-1}h_1^{-1}h_2∈HK$, so $(h_1k_1)^{-1}h_2k_2=(k_1^{-1}h_1^{-1}h_2)k_2∈HK$. By subgroup test $HK⩽G$. Suppose $(ℚ,+)$ is finitely generated, ∃ integers $n_1,…,n_r;m_1,…,m_r$ such that$$ℚ=\left<\frac{n_1}{m_1},…,\frac{n_r}{m_r}\right>$$Let $m=\operatorname{lcm}(m_1,…,m_r)$, then $ℚ=⟨\frac1m⟩$, but $\frac1{2m}∉⟨\frac1m⟩$, contradiction.