I've been trying to find <$\alpha_1$,$\alpha_2$> where $\alpha_1$,$\alpha_2$ $\in$ $\mathbb{Z}^*_{15}$ such that $\mathbb{Z}^*_{15}$=<$\alpha_1$,$\alpha_2$>. As far as I understand, the generator should be expressed in terms of the binary operation of the group. Which would mean that for $\mathbb{Z}^*_{15}$, it should be defined through multiplicative notation such that a generator a would be expressed as $<a>$:={$a^z$: z $\in$ $\mathbb{Z}$}. However, I can't find two numbers within $\mathbb{Z}^*_{15}$ that fit this criteria. The most I could find was <2,7> which generated all of $\mathbb{Z}^*_{15}$ except 11 and 14.
Additionally, the book this problem came from mentions earlier in the section that $\mathbb{Z}^*_{15}$ is not a cyclic group. Does this affect what I should be doing when looking for a generator set?
A good general method here is to use the prime factorization of $15$ to help you. We have $15=5\times 3$, and we can find primitive roots (generators of the multiplicative groups) modulo $5$ and modulo $3$. The number $2$ is actually a primitive root for both.
You can take one generator as the number in our group that's congruent to $1$ mod $5$, and congruent to a primitive root mod $3$, and as the other generator, something that's congruent to a primitive root modulo $5$, and congruent to $1$ mod $3$. These numbers would be $11$ and $7$. The first generator has order $2$, and generates a subgroup isomorphic to $\Bbb Z_3^*$ (which is trivial modulo $5$). The second one has order $4$, and generates a subgroup isomorphic to $\Bbb Z_5^*$ (which is trivial modulo $3$). Between the two of them, they generate all of $\Bbb Z_{15}^*$.