# A subgroup of the direct product of two groups

Assume that we have two groups $$G$$ and $$H$$ generated respectively by the sets $$S_G$$ and $$S_H$$. Also assume that $$S_G$$ and $$S_H$$ have the same cardinality. $$i_G$$ and $$i_H$$ are the identity elements of $$G$$ and $$H$$, while $$\circ_G$$ and $$\circ_H$$ are their respective group operators.

Consider the set $$K:=(S_G\times\{i_H\})\cup(\{i_G\}\times S_H)$$. This set, along with the component-wise operator defined as $$(a,b)\circ (c,d)=(a\circ_G c, b\circ_H d)$$, generates the direct product of the two groups $$G\times H$$.

However, this is not the only way to combine the two generating sets $$S_G$$ and $$S_H$$. Here is another approach:

Since $$|S_G|=|S_H|$$, we can build a bijection $$R:S_G\rightarrow S_H$$. Now, defining, $$S_G\star_R S_H:=\{(g,h)\mid g\in S_G \land h\in S_H \land R(g)=h\}$$ Effectively, this ‘pairs up’ elements of $$K$$, by composing them. So this new set has half the cardinality of the generating set $$K$$.

For example, suppose $$S_G=\{a,b\}$$ and $$S_H=\{x,y\}$$. In this case, $$K=\{(a,i_H),(b,i_H),(i_G,x), (i_G,y)\}$$. This set, along with the component-wise composition operator described previously, generates $$G\times H$$

If we choose the ‘pairing up’ $$S_G \star_R S_H=\{(a,x),(b,y)\}$$, we are simply taking the original generating set $$K$$ and composing the 1st and 3rd elements together to get $$(a,x)$$ and the 2nd and 4th elements together to get $$(b,y)$$

The set $$S_G\star_R S_H$$, along with the operator $$\circ$$ (as defined previously), generates a new group which we can denote as $$G \star_R H$$

$$G\star_R H$$ depends upon the chosen generating sets for each group and the bijection $$R$$ which ‘pairs up’ elements from the two generating sets.

As has been noted in the chat and the comments below, $$G\star_R H$$ is a subgroup of the direct product $$G\times H$$

Possible application

Consider the infinite dihedral group $$\mathrm{Dih}_\infty$$ and the infinite cyclic group $$\mathbb{Z}$$.

Choose the presentation $$\langle r,s\mid s^2=1,srs=r^{-1}\rangle$$ for $$\mathrm{Dih}_\infty$$ group, thus getting the generating set $$\{r,s\}$$

Choose $$\{0,1\}$$ as a generating set for $$\mathbb{Z}$$ along with addition as the group operator.

We can pair up the two generating sets to create a new set $$W=\{(r,0),(s,1)\}$$, which corresponds to a bijection $$R$$ from $$\{r,s\}$$ to $$\{0,1\}$$. The set $$W$$ generates the wallpaper group $$\mathrm{pg}$$, where the first element corresponds to the vertical translation symmetry and the second element corresponds to the horizontal glide symmetry. Thus the group $$\mathrm{pg}$$ can be shown isomorphic to $$\mathrm{Dih}_\infty\star_R\mathbb{Z}$$.

Another Example

We have chosen $$\mathrm{K4}$$ (Klein-4 group) and $$\mathrm{D3}$$ (symmetry group of equilateral triangle) to demonstrate $$\mathrm{K4}\star_R\mathrm{D3}$$

Chosen presentations:

For $$\mathrm{K4}$$, from here: $$\langle x,y \mid x^2 = y^2 = (xy)^2 = 1 \rangle$$

For $$\mathrm{D3}$$, from here: $$\langle a,b \mid a^3= b^2 = 1, ab = ba^{-1} \rangle$$

Thereby we get the generating sets $$\{x,y\}$$ and $$\{a,b\}$$ respectively.

We choose the following pairing, to get a new generating set $$\{(x,a),(y,b)\}$$, and its corresponding bijection $$R$$. We can use this generating set, along with the composition operator as described previously to get $$\mathrm{K4}\star_R\mathrm{D3}$$

As we can see, $$\mathrm{K4}\star_R\mathrm{D3}$$ is isomorphic to $$\mathrm{D6}$$, a subgroup of direct product $$\mathrm{K4}\times\mathrm{D3}$$

Edit:

Discussed further in this chat

Edit #2:

I have tried to clarify my original question, but there may still be issues that I have overlooked. All suggestions on how to improve this post are welcome.

• It looks a bit like an amalgamated free product, but of a direct product. I doubt it's new. Commented Jul 31 at 10:32
• Also, the construction is not at all clear. Commented Jul 31 at 10:35
• If I understand correctly, the elements of $\mathrm{Dih}_\infty\star\mathbb{Z}$ will look like $(g,n)$ with $g\in \mathrm{Dih}_\infty$ and $n\in \Bbb Z$. And the operation is done component-wise. So, isn't this just $\mathrm{Dih}_\infty\times\mathbb{Z}$? Commented Jul 31 at 10:59
• The RHS of $$S_G\star S_H:=\{(g,h)\mid g\in S_G \land h\in S_H \land R(g)=h\}$$ depends on $R$ when the LHS does not. Commented Jul 31 at 11:35
• Isn't this just a subgroup of the direct product? If so I don't think it's a 'A new type of group product' or needs new notation, but I think it can be studied. Commented Jul 31 at 15:25