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 $X$ be the set of generators. We define the free group $F(X)$ on $X$ to be the set of reduced words in the alphabet $\left\{x_i, x_i^{-1}: x_i ∈ X\right\}$. (Reduced means that there are no adjacent $x x^{-1}$ or $x^{-1} x$ pairs). The empty word gives the identity. The group operation is concatenation (and reduction if necessary).
         Now let $R$ denote the set of relations, which are elements of $F(X)$. We let $⟨⟨B⟩⟩$ denote the normal subgroup of $F(X)$ generated by $B$ (the intersection of all normal subgroups containing $B)$
         The group $⟨X ∣ R⟩$ given by generators $X$ and relations $R$ is now defined to be $F(X) /⟨⟨R⟩⟩$.
      2. As $b^2=1$ we have $b$ is self-inverse, so the third relation may be written as $$ a b=b a^{-1}  (\text{and hence } a^{-1} b=b a) $$ Every element of $D_{2 n}$ is a word in $a, a^{-1}$ and $b$, and using the above equation we may systematically rewrite it so that every $b$ comes before every $a$ or $a^{-1}$. As $b^2=1$ and $a^n=1$ this gives us exactly the $2 n$ elements listed.
         An element $x$ is central if and only if it commutes with the generators $a, b$. Now $a^j b=b a^{-j}$ inductively from above, so $a^j$ is central if and only if $a^{2 j}=1$, ie $j=\frac{n}{2}$. Moreover $a\left(b a^i\right)=b a^{i-1}$ while $\left(b a^i\right) a=b a^{i+1}$ and these are distinct for $n>2$. If $n=2$, of course, $a=a^{-1}$ and $b a b^{-1}=a$ so the group is Abelian.
         So the centre is
         1) trivial if $n$ is odd,
         2) $\left\{1, a^m\right\}$ if $n=2 m>2$
         3) the whole group if $n=2$.
   2. We compute commutators as follows: \begin{gathered} {\left[b a^i, a^j\right]=b a^i a^j a^{-i} b a^{-j}=b a^j b a^{-j}=b^2 a^{-2 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^{j-2 i} a^j=a^{2(j-i)},} \\ {\left[a^i, a^j\right]=1 .} \end{gathered} so the commutator $D_{2n}^{(2)}=\left< a^2\right>$ is Abelian and $D_{2n}^{(n)}=1$ for $n>2$.

      Theorem 67 $G$ is solvable iff the derived length of $G$ is finite, i.e. there exists $k∈ℕ$ such that $G^{(k)}=\{e\}$;

   3. 1. $(D_6)_2=\left< a^2\right>=\left\{1, a, a^2\right\}$. Now further commutation with $b a^i$ sends $a ↦ a^{-1}=a^2$ and $a^2 ↦ a^{-2}=a$, so in fact we get $(D_6)_n=(D_6)_2$ for $n≥2$.
      2. \begin{align*} \left[a^i, a^{2 j}\right]&=1\\ \left[b a^i, a^{2 j}\right]&=a^{-4 j} \end{align*} We know $(D_{2n})_2=\left[D_{2 n}, D_{2 n}\right]=\left< a^2\right>$ and inductively $(D_{2n})_m=\left< a^{2^{m-1}}\right>$
         So $(D_{2n})_m=\{1\}$ for some $m$ iff $o(a)=n$ divides $2^{m-1}$, ie iff $n$ is a power of 2.
      3. We observe that $$ g[x, y] g^{-1}=g x y x^{-1} y^{-1} g^{-1}=\left(g x g^{-1}\right)\left(g y g^{-1}\right)\left(g x^{-1} g^{-1}\right)\left(g y^{-1} g^{-1}\right)=\left[g x g^{-1}, g y g^{-1}\right] $$ so $[G, G]$ is a normal subgroup of $G$. Now let $G$ be any nonabelian simple group, eg. $A_n$ for $n ≥ 5$. Then $[G, G]$ is a nontrivial normal subgroup so equals $G$, and $G_i=G$ for all $i$.