2. 1. 1. Explain what it means for a group to be defined in terms of generators and relations.
      2. Let a group be defined as: $$ D_{2 n}=\left< a, b ∣ a^n=b^2=1, b a b^{-1}=a^{-1}\right> $$ where $n>1$.
         Prove that $D_{2 n}$ consists of
         1) elements $a^i: i=0,1, …, n-1$, together with
         2) elements $b a^i: i=0,1, …, n-1$.
         Furthermore determine the centre $$ Z\left(D_{2 n}\right)=\left\{x∈D_{2 n}: x y=y x \text { for all } y∈D_{2 n}\right\} $$
   2. The commutator subgroup $[G, G]$ of a group $G$ is defined to be the subgroup generated by all elements $x y x^{-1} y^{-1}$ where $x, y∈G$.
      1. Find the commutator subgroup of $D_{2 n}$.
      2. For a group $G$, define a sequence of subgroups of $G$ by $$ G^{(1)}=G, \text { and } G^{(i)}=\left[G^{(i-1)}, G^{(i-1)}\right] \text { for } i>1 . $$ Determine $\left(D_{2 n}\right)^{(i)}$ for all $i$.
   3. For a group $G$, define a sequence of subgroups of $G$ by $$ G_1=G, \text { and } G_i=\left[G, G_{i-1}\right] \text { for } i>1 $$ where $[G, H]$ denotes the subgroup generated by all elements $x y x^{-1} y^{-1}$ for $x∈G$ and $y∈H$.
      1. Determine $\left(D_6\right)_i$ for all $i ⩾ 1$.
      2. For which $n$ is $\left(D_{2 n}\right)_i=\{1\}$ for some $i$ ?
      3. Give an example, with brief justification, of a nontrivial group $G$ for which $G_i=G$ for all $i$.

Solution

2. 1. 1. Let $F(X)$ be free group on a set $X$, let $R$ be a collection of elements of $F(X)$, let $⟨⟨R⟩⟩$ be the normal closure of $R$. Then the group with presentation $⟨X|R⟩$ is $F(X)/⟨⟨R⟩⟩$.
      2. Using $ba=a^{-1}b$ move every $a$ after $b$, so $D_{2n}=\{b^ia^j|i,j∈ℤ\}$.
         Using $a^n=b^2=1$, restrict to $i∈\{0,1\},j∈\{0,1,…,n-1\}$, so $|D_{2n}|$ ≤ 2n.
         If $b^{i_1}a^{j_1}=b^{i_2}a^{j_2}$, then $b^{i_1-i_2}=a^{j_2-j_1}⇒2|i_1-i_2,n|j_2-j_1$, so $|D_{2n}|$ = 2n.
         For $n=2$, $ab=ba$, so $D_4$ is abelian, $Z(D_4)=D_4$.
         For $n>2$, $(ba^j)a=baa^j=a^{-1}(ba^j)≠a(ba^j)$, so $ba^j∉Z(D_{2n})$, so $Z(D_{2n})⊂⟨a⟩$.
         $⟨a⟩$ is abelian, so $a^j$ is in the center iff it commutes with $b$.
         $$a^jb=ba^j=a^{-j}b⇒n|j-(-j)$$ If $n$ is even, $j=0,n/2$ satisfy $n|2j$, so $Z(D_{2n})=\{1,a^{n/2}\}$.
         If $n$ is odd, only $j=0$ satisfy $n|2j$, so $Z(D_{2n})=\{1\}$.
         
   2. 1. For $n=2$, $D_4$ is abelian, $x y x^{-1} y^{-1}=1$, so $[D_4,D_4]=\{1\}$. For $n>2$, \begin{align*} \left[a^i, a^j\right]&=1\\ \left[b a^i, a^j\right]&=\left[a^j,b a^i\right]^{-1}=b a^i a^j a^{-i} b a^{-j}=b a^j b a^{-j}=a^{-j}a^{-j}=a^{-2 j}\\ \left[b a^i, b a^j\right]&=b a^i b a^j a^{-i} b a^{-j} b=a^{-i}a^ja^{-i} a^j=a^{2(j-i)} \end{align*} these elements generate $D_{2n}^{(2)}=⟨a^2⟩$.
      2. If $n$ is odd, $D_{2n}^{(2)}=⟨a^2⟩=⟨1,a^2,a^4,…,a^{n-1},a^{n+1},…,a^{2n-2}⟩=⟨1,a,a^2,…,a^{n-1}⟩≅C_n$.
         If $n$ is even, $D_{2n}^{(2)}=⟨a^2⟩=⟨1,a^2,a^4,…,a^{n-2}⟩≅C_{n/2}$.
         Since $D_{2n}^{(2)}$ is abelian, for $i>2$ we have $D_{2n}^{(i)}=\{1\}$.
   3. 1. Since $a^3=1$, $(D_6)_2=[D_6,D_6]=⟨a^2⟩=⟨a⟩$. \begin{align*} \left[a^i, a^j\right]&=1\\ \left[b a^i, a^j\right]&=a^{-2 j}=a^j \end{align*} So $(D_6)_3=[D_6,⟨a⟩]=⟨a⟩$. Similarly $(D_6)_i=⟨a⟩$ for all $i≥2$.
      2. $[b, a^j]=a^{-2 j}$, so $(D_{2n})_2=⟨a^2⟩$, inductively $(D_{2n})_{i+1}=⟨a^{2^i}⟩$ for $i≥1$.
         If $n=2^k$ for some $k≥1$, then $(D_{2n})_{k+1}=⟨a^{2^k}⟩=\{1\}$.
         If $n$ is not a power of 2, then $n=(2k+1)2^l$ for some $k≥1,l≥0$. $$(D_{2n})_{l+1}=⟨a^{2^l}⟩=\{1,a^{2^l},…,a^{2k⋅2^l}\}$$ For $i≥l$, since $n|(2k+1)2^i$,$$a^{-2^i}=a^{2k2^i}∈⟨a^{2^{i+1}}⟩⇒(D_{2n})_{i+1}⊂(D_{2n})_{i+2}$$on the other hand$$a^{2^{i+1}}=(a^{2^i})^{2^i}∈⟨a^{2^i}⟩⇒(D_{2n})_{i+2}⊂(D_{2n})_{i+1}$$so $(D_{2n})_{i+1}=(D_{2n})_{i+2}$. So $(D_{2n})_{i+1}=(D_{2n})_{l+1}≠\{1\}$ for all $i≥l$.
      3. $G_i=G$ for all $i⇔[G,G]=G$. By Lemma 64(i) $[G,G]⊲G$.
         For $G=A_5$, since $G$ is nonabelian, $[G,G]$ is nontrivial, also $G$ is simple, so $[G,G]=G$.