Let G = $\displaystyle {(\mathbb{Z}/7\mathbb{Z})^*} \times {{(\mathbb{Z}/11\mathbb{Z})^*} $. How many elements are there in the group G? Also show that G is not cyclic.

I know that no. of elements in $\displaystyle {(\mathbb{Z}/7\mathbb{Z})^*}$ = 6 and no. of elements in $\displaystyle {(\mathbb{Z}/11\mathbb{Z})^*}$ = 10. So can i conclude that the total number of elements is then 60?

As for the cyclic part, i dont have any idea where to start from. I know that a group is cyclic if for every element $\displaystyle g \in G$, $\displaystyle g=r^i$, where r is the generator of the cyclic group. But how do i show that such a generator does not exist?

Thanks.