Any mathematical software to verify no unique products.

Question

Let $D^3= D \times D \times D$ where $D = D_\infty$ where we see $D$ as the group generated by $\mathbb{Z}$ and element $0^*$ of order $2$ such that $0^*n0^*=-n$ for all $n \in \mathbb{Z}$. Letting $n0^*$ be $n^*$ gives the following multiplication: $$nk=n+k \\ nk^*=(n+k)^* \\ n^*k=(n-k)^* \\ n^*k^*=(n-k)$$
Now let $G$ be the subgroup of $D^3$ generated by $a=(1,0^*,0^*)$ and $b=(0^*,1,1^*)$.

Let $S$ be the set of $14$ elements i.e. $S=$$\left\{ (0,0,2),(0,0,-2),(2^*,1,1^*),(2^*,-1,-1^*),(0^*,1,-1^*),(0^*,-1,1^*),(0^*,-1,-1^*),(1,2^*,0^*),(-1,2^*,0^*),(1,0^*,-2^*),(-1,0^*,2^*),(1,0^*,0^*),(-1,0^*,0^*) \right\} \subset G $

Now $S.S$ consists of $196$ elements and I want to show that every element of $S.S$ also have another representation (i.e. non unique products) as $(a_1,b_1,c_1)(a_2,b_2,c_2)$ where both $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are of the form $$E_0: (\text{even}, \text{even},\text{even}) \\ E_1 : (\text{even}^*,\text{odd}, \text{even}^*) \\ E_2 : (\text{odd}, \text{even}^*,\text{even}^*) \\ E_3: (\text{odd}^*,\text{odd}^*,\text{odd})$$ which are basically just the types of elements $G$ consists of.

One way to check it is by hand.

I want to know can we check it by some mathematical software or by writing some programme (which I am a dud in). If we can, can somebody help me here, I checked a few with hands, but the list is long and it will be tiresome to check all $196$.

Answer 1

Briefly the idea of a python program:

• we form list $L$ of $E_1,\,E_2,\,E_3,\,E_4$ where odd=[1,-1,3,-3] and even=[0,2,-2,4,-4]
• then we compute $xy$ for $(x,y)$ in $L^2$ and store unique products in a dictionary d
• then we compute $xy$ for $(x,y)$ in $S^2$ and check whether it is in d

The code is here http://pastebin.com/KzEYF6Jr and the result with $(0^*,1,1^*)$ included is here http://pastebin.com/91mw6Ms5 .

P.S. Please don't accept this too soon as I want to know if there is a software, which is a fishing-rod, but not the fish.