Let $K⊴G$ and let $\bar{H}≤G/K$. Let $π:G→G/K$ denote the quotient map $g↦g K$. Show that$$H=π^{-1}(\bar{H})=\{g∈G:gK∈\bar{H}\}$$is a subgroup of $G$, containing $K$ as a normal subgroup, with $H/K=\bar{H}$. Show further that if $\bar{H}⊴G/K$ then $H⊴G$.
subgroup contains normal subgroup as normal subgroup and is also normal
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.
The dihedral group $D_{2 n}$ has presentation$$\left<a,b\middle|a^n=b^2=1,bab^{-1}=a^{-1}\right>$$Verify that this group has $2 n$ elements, all of the form $a^i$ or $b a^i$, and that $\left(b a^i\right)^2=1$. Interpret this geometrically.
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).
Identify the following groups from their presentation
$(ℤ⊕ℤ)/⟨(n,m)⟩≅ℤ⊕ℤ/⟨d⟩,d=\gcd(n,m)$. Describing groups with given presentation? Group Isomorphic to (ℤ⊕ℤ)/⟨(4,2)⟩ How to describe the quotient group (ℤ⊕ℤ)/⟨(4,-6)⟩
Let $G=\left<x, y∣x^2=y^2=1\right>$.
Let $G$ be a non-Abelian group of order 8.Show that one of the non-Abelian groups may be identified with the quaternion group$$Q_8=\{±1,±i,±j,±k\}$$where we have the usual quaternionic relations$$i^2=j^2=k^2=-1; i j=k=-j i .$$
Write down all composition series of the following groups and verify the Jordan-Hölder Theorem for them:$$C_{18},D_{10},D_8,Q_8.$$
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: $\{e\}◃C_2◃C_4◃C_4×C_2$ $\{e\}◃C_2◃C_2×C_2◃C_4×C_2$
Let $H$ and $K$ be subgroups of a group $G$. Show that$$HK=\{h k: h∈H, k∈K\}$$is a subgroup of $G$ if and only if $H K=K H$.
(⇒) 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 $HK⩽G$.
Show that $(ℚ,+)$ is not finitely generated.
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.