Solution:

since, $G_1=[1,2,3,4,5,6] \times 7 \quad[Given]$

since $a \in G$ belongs for every element $x \in G$.

Let's check for 1 -

$ \begin{aligned}\\ &1^1=1 \\\\ &1^2=1 \\\\ &1^3=1 \\\\ &1^4=1 \\\\ &1^5=1 \\\\ &1^6=1\\ \end{aligned}\\ $

that is wrong.

For 2:

$ \begin{aligned}\\ &2^{\prime}=2 \\\\ &2^2=4 \\\\ &2^3=8 \% …