Prove that the following groups are isomorphic

abstract-algebracombinatorial-group-theoryfree-productgroup-isomorphismgroup-theory

Consider the following groups.

  1. $\langle a,b,c~ |~aba^{-1}b^{-1},aca^{-1}c^{-1}\rangle$

  2. $(\Bbb Z * \Bbb Z) \times \Bbb Z$

  3. $((\Bbb Z \times \Bbb Z)*(\Bbb Z \times \Bbb Z))/N$, where $N$ is the normal subgroup generated by the element $(1,0)_1*(0,-1)_2$ (the subscript indicates that, for example, $(1,0)_1$ belongs to the first $\Bbb Z \times \Bbb Z$ factor).

I know that to show two groups are isomorphic we usually use the First Isomorphism Theorem, but I am not familiar with free products, so I got stuck. How do I have to proceed? Any hints will be appreciated.

Best Answer

First, consider the following group with labeled generators

$$G = \underbrace{(\mathbb{Z}\ast\mathbb{Z})}_{(~b~~~\!,\!~~~c~)}\times\underbrace{\mathbb{Z}}_{a}$$

So we can start out with $G = \langle a,b,c\rangle$. Now, let's find relations between these generators. We know that $b$ and $c$ have no relations between them due to the free product. Moreover, by definition of the direct product we know that the only relations involving $a$ are $ab=ba$ and $ac=ca$. Thus, we have

$$(\mathbb{Z}\ast\mathbb{Z})\times\mathbb{Z} = \langle a,b,c~|~ aba^{-1}b^{-1}=aca^{-1}c^{-1}=1\rangle.$$

To find a presentation of $(\mathbb{Z}\times\mathbb{Z})\ast(\mathbb{Z}\times\mathbb{Z})/N$, we can follow the same procedure, with one extra step. Consider the following group with labeled generators:

$$G = \underbrace{(\mathbb{Z}\times\mathbb{Z})}_{(~a~~~\!,\!~~~b~)}\ast\underbrace{(\mathbb{Z}\times\mathbb{Z})}_{(~c~~~\!,\!~~~d~)}$$

where $N = \langle ad^{-1}\rangle$. So, we can start out with $G = \langle a,b,c,d\rangle$. Now, let's find relations among these generators in $G$. Since we just have direct products and free products, the only relations we need are $ab=ba$ and $cd=dc$. When we quotient $G$ by $N$, we simply add the relation $ad^{-1}=1$. Thus, we get

$$G/N = \langle a,b,c,d~|~aba^{-1}b^{-1}=dcd^{-1}c^{-1}=ad^{-1}=1\rangle.$$

Since $ad^{-1}=1$ is equivalent to $a=d$, we can replace $d$ with $a$ and get the presentation

$$G/N = \langle a,b,c~|~ aba^{-1}b^{-1}=aca^{-1}c^{-1}=1\rangle.$$

Therefore, both the second and the third group have the presentation of the first group, and so they are all isomorphic.

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