Example 23.1 (Presentation of some groups):

We describe some of the commonly occurring groups in terms of generators and relations. Some of these would appear as fundamental groups of spaces that we have already encountered or would do so in the next few lectures.

1. If we take the free group on two generators $a, b$ and take $H = \mathbb{Z}\times \mathbb{Z}$ then every commutator $a^mb^na^{-m}b^{-n}$ is a relator and hence each of the equations $a^mb^na^{-m}b^{-n} = 1$ is a relation. However, all of them may be derived from the single relation $aba^{-1}b^{-1} = 1$. For example, we derive the relation $a^2ba^{-2}b^{-1} = 1$ as follows
   $$a^2ba^{-2}b^{-1} = a(aba^{-1}b^{-1})ba^{-1}b^{-1} = aba^{-1}b^{-1} = 1.$$
   Thus $\mathbb{Z}\times \mathbb{Z}$ has presentation
   $$\mathbb{Z}\times \mathbb{Z} = \langle a, b\;\vert\; ab = ba\rangle \eqno(23.5)$$

2. The cyclic group of order $n$ has presentation
   $$\mathbb{Z}_n = \langle a\;\vert\; a^n = 1\rangle \eqno(23.6)$$

3. Recall from lecture 20 that the fundamental group of the Klein's bottle is given by the presentation
   $$\mathbb{Z}\ltimes \mathbb{Z} = \langle a, b\;\vert\; aba = b\rangle \eqno(23.7)$$

4. This example is from [15], p. [?]. Let us consider the group $G$ given by the presentation
   $$G = \langle a, b\;\vert\; a^2 = b^4 = 1,\; bab = a\rangle \eqno(23.8)$$
   To understand this group concretely, let us derive some consequences of the three displayed relations. Multiplying $bab = a$ on the left/right by $a$ gives the relations $(ab)^2 = 1$ and $(ba)^2 = 1$. Further,
   $$ab^3 = (ab)b^2 = b^3(bab)b^2 = b^3ab^2 = ba.$$
   We conclude from this that $G$ consists of the elements
   $$\{1, a, b, b^2, b^3, ab, ba, ab^2\} \eqno(23.9)$$
   This however does not preclude further simplifications to a group of smaller order though it seems unlikely. The group has atleast three elements of order two and so if the elements listed in (21.9) are distinct then $G$ must be the dihedral group $D_4$ of order eight if it is non-abelian or else must be an abelian group. In any case there must be atleast five elements of order two (why?). It is easy to see that $ab^2$ has order two. The map $f:{a, b}\longrightarrow D_4$ given by
   $$f(a) = (13),\;f(b) = (1234)$$
   respects the given relations since $(13)^2 = 1$, $(1234)^4 = 1$ and $(1234)(13)(1234) = (13)$. Hence $f$ extends to a surjective group homomorphism $f:F_2\longrightarrow D_4$. Since the kernel contains $a^2, b^4$ and $bab$ we get a surjective group homomorphism $G\longrightarrow D_4$ and we conclude that $G$ is indeed $D_4$.