Statement A is false.
$Z^{\ast}_{8}$ is not even a group.
Statement C is false. We know that there is a non-abelian group G of order $6.$ Every cyclic group is abelian So, every non-abelian group is non-cyclic. Now, for this non-cyclic group G of order $6,$ every proper subgroup will have order $1$ or $2$ or $3,$ which are all cyclic groups.
Option B:
It’s not true because $< 49 >= Z_{60}$ and $49$ is not prime. Also $< 1 >= Z_{60}$ and $1$ is not prime either.
Detailed Explanation:
We know the inverse of any element in $Z_n$ given as following:
Theorem $6.9.$ The order of an element $a \in \mathbb{Z}_{n}$ (under addition modulo $n$ ) is $n / \operatorname{gcd}(a, n)$.
The general formula for the order of $x$ in $\mathbb{Z}_{n}$ is $\frac{n}{\operatorname{gcd}(x, n)}$.
So, in $Z_{n}$ an element is a generator if and only if it has an order $60.$
If an element has order $60,$ then it means it generates 60 elements, hence, it is a generator.
So, in $Z_{n}$ an element $x$ is a generator if and only if $\operatorname{gcd}(x, n)=1$.
Option D is true. Elements $1,3,5,7$ have inverse.
We know that in multiplicative modulo $n$, an element $x$ has multiplicative inverse iff $\mathrm{x}, \mathrm{n}$ are co-prime i.e. $\operatorname{gcd}(\mathrm{x}, \mathrm{n})=1$.
