Seeing is believing

Note: We use the convention that $1$ is the identity/neutral element of some group, and abbreviate some binary operation $*$ between elements $x, y \in G$ , $G$ a group, as $*(x, y) = x*y = xy$ .

Consider the dihedral group. A geometric or visual interpretation deals with polygons, which are 2 dimensional. One example of a dihedral group is to consider all vertices of a triangle under the binary operation of moving a step forward, and each vertex has some orientation (which we will denote by the presence or absence of $\tau$ ). Label the vertices $1, \sigma, \sigma^2$ . $1$ means you didn't move forward at all (stayed at the same point), $\sigma$ means you moved a step forward, and $\sigma^2$ means you moved two steps forward. But, if you multiply each vertex by $\tau$ , it's almost like adding a negative sign to $\sigma$ ; $\tau \sigma$ means we first move from $1$ to $\sigma$ , then change the direction of $\sigma$ in such a way that if we then try to do $\sigma$ again, meaning $\sigma (\tau \sigma )$ , we end up back where we started, namely at $1$ , except with orientation $\tau$ . In other words, $\tau \sigma = \sigma^{-1} \tau$ . It is then apparent that there are $6$ unique elements in this dihedral group: $1, \sigma, \sigma^2, \tau, \tau \sigma, \tau \sigma^2$ .

Now consider the dihedral-like group which consists of the vertices of a dodecagon, $1, \sigma, \sigma^2, ..., \sigma^{11}$ , and each vertex has some orientation, which is described by the presence or absence of any or all of $2018$ distinct elements, denoted $\tau_1, \tau_2, ... , \tau_{2018}$ which under the operation of "moves forward", for each $i \in \mathbb{N} : 1 \leqslant i < 2018$ , each $\tau_i$ , behaves just like $\tau$ from the previously mentioned dihedral group, and for $1 \leqslant i < j \leqslant 2018$ , $i,j \in \mathbb{N}$ , $\tau_i \tau_j = \tau_j \tau_i$ . How many distinct elements are in this dihedral-like group? If the answer can be represented in the form $3 \cdot 2^a$ , find $a$ .

Bonus : To be fully rigorous, you should ignore the geometric intuition (for a minute--use it to give you an idea how to solve the problem but do not rely on it and be rigorous algebraically) and try to formulate the problem more precisely, only thinking algebraically, i.e. using bracket notation. You could define dihedral-like group as $G = \left \langle \text{set of generators} | \text{ relations between generators} \right \rangle$ , which is very precise. Also, without geometric intuition, if we define the finite group $G'$ as having order $n$ as having elements $1, \sigma, \sigma^2, \sigma^3, ..., \sigma^{n-1}$ , how do you know that $1, \sigma, \sigma^2, ..., \sigma^{n-1}$ are all distinct? Also, how do we know that the dihedral-like group I discussed is actually a group in the first place? You should prove that.