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 \% 7=1 \\\\ &2^4=16 \% 7=2 \\\\ &2^5=32 \% 7=4 \\\\ &2^6=64 \% 7=1\\ \end{aligned}\\ $

that is wrong.

For 3:

$ \begin{aligned}\\ &3^1=3 \\\\ &3^2=9 \% 7=2 \\\\ &3^3=27 \% 7=6 \\\\ &3^4=81 \% 7=4 \\\\ &3^5=243 \% 7=5 \\\\ &3^6=729 \% 7=1\\ \end{aligned}\\ $

that is right.

For 4:

$ \begin{aligned}\\ &4^{\prime}=4 \\\\ &4^2=16 \% \cdot 7=2 \\\\ &4^3=64 \% 7=1 \\\\ &4^4=256 \% 7=4 \\\\ &4^5=1024 \% 7=2 \\\\ &4^6=4096 \% 7=1\\ \end{aligned}\\ $

that is wrong.

For 5̤:

$ \begin{aligned}\\ &5^1=5 \\\\ &5^2=25 \% 7=4 \\\\ &5^3=125 \% 7=6 \\\\ &5^4=625 \% 7=2 \\\\ &5^5=3125 \% 7=3 \\\\ &5^6=15625 \% 7=1\\ \end{aligned}\\ $

that is right.

For 6:

$ \begin{aligned}\\ &6^{\prime}=6\\\\ &6^2=36 \% 7=1\\\\ &6^3=216 \% 7=6\\\\ &6^4=1296 \%=1\\\\ &6^5=7776 \%=6\\\\ &6^6=46656 \% 1.7=1\\ \end{aligned}\\ $

Therefore, 3 and 5 are generators.. Hence, G is a cyclic group with generators 3 and 5 .