Consider $G\_n$ as the multiplicative cyclic group given by the $n^{th}$ roots of unity.

$$G\_n = \left\{ e^{ 2ik\pi/n} \mid 1\leq k \leq n \right\}$$

Now construct a sequence from each $G\_n$ by $(e^{2i1\pi/n}, e^{2i2\pi/n},\dots, e^{2ik\pi/n},\dots,e^{2in\pi/n}, 0, 0, 0,\dots$).

Now consider the set S of all such sequences, 
$$S=\left\{(e^{2i1\pi/n}, e^{2i2\pi/n},\dots, e^{2ik\pi/n},\dots,e^{2in\pi/n}, 0, 0, 0,\dots) : 1\leq n \lt \infty\right\}$$
Will I be right in concluding that no matter what operation I impose on $S$, the structure of $S$ will never be a semi-group/vector space/normed vector space? 

I wondered about the practical utility of creating $S$. Any help?

An observation is that $(1,0,0,\dots)$ is the only limit (obtained by pointwise convergence of first sequence) contained by $S$ (when $S$ is viewed as a subset of $C^n$.
 It contains no other elements that are obtained by pointwise convergence of each sequence. The pointwise limit of each sequence is of the form $(1,1,....,\underbrace{1}\_{n^\text{th}\text{ place}},0,0,\dots)$.