# Group Presentation of the Direct Product.

This is Exercise 1.2.5 and Exercise 1.2.6 of "Combinatorial Group Theory: Presentations of Groups in Terms of Generators and Relations," by Magnus et al.

## The Details:

Definition 1: Let $$\langle a, b ,c, \dots \mid P, Q, R, \dots \rangle$$ be a group presentation, where $P, Q, R, \dots$ are relators, not relations (i.e., words, not equations). We say the words $W_1$ and $W_2$ in $a, b, c, \dots$ are equivalent, denoted $W_1\sim W_2$, if the following operations applied a finite number of times, change $W_1$ into $W_2$:

(i) Insertion of one of the words $P, P^{-1}, Q, Q^{-1}, R, R^{-1}, \dots$ or one of the trivial relators between any two consecutive symbols of $W_1$, or before $W_1$, or after $W_1$.

(ii) Deletion of one of the words $P, P^{-1}, Q, Q^{-1}, R, R^{-1}, \dots$ or one of the trivial relators, if it forms a block of consecutive symbols in $W_1$.

## The Question(s):

Exercise 1.2.5 Suppose $G=\langle a, b\mid P(a, b), Q(a, b)\rangle$ and $H=\langle x, y\mid S(x, y), T(x, y)\rangle$. Then show that the direct product $G\times H$ has the presentation $$\langle a, b, x, y\mid P(a, b), Q(a, b), S(x, y), T(x, y), ax=xa, ay=ya, bx=xb, by=yb\rangle.$$

[$\color{red}{\text{Hint}}$: If $G$ is presented under the mapping $\theta: a\mapsto g, b\mapsto g'$, and $H$ is presented under the mapping $\phi: x\mapsto h, y\mapsto h'$, then show that the combined mapping $\theta\times \phi:a\mapsto (g, 1), b\mapsto (g', 1), x\mapsto (1, h), b\mapsto (1, h')$ determines a homomorphism of the alleged presentation for $G\times H$ onto $G\times H$. Next show that each element of the alleged presentation can be defined by a word $U(a, b)V(x, y)$. Show that if $U(a, b)V(x, y)\sim U'(a, b)V'(x, y)$, then $U(g, g')=U'(g, g')$ and $V(h, h')=V'(h, h')$ by mapping the alleged presentation for $G\times H$ into $G$ under $a\mapsto g, b\mapsto g', x\mapsto 1, y\mapsto 1$, and into $H$ under $a\mapsto 1, b\mapsto 1, x\mapsto h, y\mapsto h'$.]

Exercise 1.2.6: Generalise Exercise 1.2.5 to arbitrary presentations $G$ and $H$. Generalise Exercise 1.2.5 to an arbitrary number of groups $G, H, \dots$.

## My Attempt:

The hint for Exercise 1.2.5 is very detailed. The map $\theta\times\phi$ clearly defines a homomorphism from the presentation to $G\times H$ because on the generators, for example, \begin{align}(\theta\times\phi)(ab)&=(gg', 1) \\ &=(g\times g', 1\times 1) \\ &=(g, 1)\times (g', 1) \\ &=(\theta\times\phi)(a)(\theta\times\phi)(b). \end{align}

That the generators of $G$ commute with the generators of $H$ in the alleged presentation for $G\times H$ means we can clearly move blocks in $\{a, b\}$ in a word past and to the left of $x, y$, resulting in a word of the form $U(a, b)V(x, y)$.

I have no idea how to show that if $U(a, b)V(x, y)\sim U'(a, b)V'(x, y)$, then $U(g, g')=U'(g, g')$ and $V(h, h')=V'(h, h')$.

As for Exercise 1.2.6, I have found that, for $G=\langle \mathcal G_G\mid \mathcal R_G\rangle$ and $H=\langle \mathcal G_H\mid \mathcal R_H\rangle$, $G\times H$ has the presentation

$$\langle \mathcal G_G\cup \mathcal G_H\mid \mathcal R_G\cup\mathcal R_H\cup\{xy=yx : x\in \mathcal G_G, y\in \mathcal G_H\}\rangle,$$

though I don't know how to prove it.

I have no idea how to generalise it to an arbitrary number of groups.

• An idea to prove your idea for 1.2.6: if $x_1 y_1....x_n y_n$ is always $1$ in $G\times H$ (seen as a quotient of the free group on $\mathcal{G}_G\cup \mathcal{G}_H$, which one can clearly do), then you can use the commutativity condition to get that $x_1^{k_1}...x_m^{k_m} = y_1^{l_1}...y_p^{k_p}$ . But then the left side is in $G$, the right one is in $H$: thus both sides are equal to $1$, and these are relations in $\mathcal{G}_G, \mathcal{G}_H$. Therefore the only relations are the ones in $G$, the ones in $H$, and the commutativity condition (obviously, these ones are relations). – Maxime Ramzi Jul 30 '17 at 19:58
• My previous comment is not really rigorous, and I'll let you detail the notations etc. – Maxime Ramzi Jul 30 '17 at 19:59
• @Shaun hello sir have you got the solution of this question? Please tell me – MANI Nov 29 '19 at 16:18
• @Shaun actually sir I was thinking about the comment given by Derek holt at this question math.stackexchange.com/questions/2319583/… – MANI Nov 29 '19 at 16:33
• @Shaun my attempt: since $U(a,b)V(x,y)\leftrightsquigarrow U'(a,b)V'(x,y)$ therefore there exists a finite sequences of relaters such that $\theta \times \phi(U(a,b)V(x,y)) =\theta \times \phi(U'(a,b)V'(x,y))$. Therefore $(U(g,g'),V(h,h'))=(U'(g,g'),V'(h,h'))$. On comparing components we have the result. Is this seems correct? – MANI Nov 29 '19 at 19:02

I am trying to give a proof of $$1.2.6$$ using free product of groups
Let $$G_1\cong\frac{F_1}{R_1}$$ and $$G_2\cong\frac{F_2}{R_2}$$ be presentations of two groups, where $$\mid F_i\mid=a_i$$(may be finite or infinite) with isomorphism $$\phi_n,~n=1,2$$. Then we claim that the the following sequence is exact:
$$1 \longrightarrow \langle R_1,R_2, [F_1,F_2]\rangle \longrightarrow F_1*F_2 \longrightarrow G_1\times G_2\longrightarrow 1$$
The universal property of free product of groups gives the map $$\phi: F_1*F_2\to G_1\times G_2$$ defined as $$\phi \mid _{F_n}=i_n\phi_n$$, where $$i_n:G_n\to G_1\times G_2$$ is usual inclusion map. Since the images of $$\phi_n$$ commutes in $$G_1\times G_2$$. Therefore $$\langle R_1,R_2,[F_1,F_2]\rangle \subseteq Ker(\phi)$$ similarly since $$\phi$$ factors through $$\phi_1$$ and $$\phi_2$$ so $$Ker(\phi)\subseteq\langle R_1,R_2,[F_1,F_2]\rangle$$ .