This is one of my assignment using fundamental theorem of finitely generated abelian groups. However, I don't really know how to find the smallest generating sets of ${(\mathbb{Z}/_{51})}^\times$ and ${(\mathbb{Z}/_{15})}^\times\oplus{(\mathbb{Z}/_{5})}^\times$ without trying all of the element.
Note: I already solved it without the above theorem. Here is my proof:
Let $A = {(\mathbb{Z}/_{15})}^\times\oplus{(\mathbb{Z}/_{5})}^\times$ and $B = {(\mathbb{Z}/_{51})}^\times$
Assume there exists group isomorphism $f: A$ → $B$. Then since $a^8=1,\forall a\in A$, we have $f(a)^8=f(a^8)=f(1)=1$
Because f is isomorphism, $b^8=1,\forall b\in B$. This is impossible since $2^8=13\ne 1$.
Therefore, no isomorphism exists from A to B (Q.E.D)
Edit: I miscalculate the order of 2 and it should be 7 (which $7^8\ne 1$), sorry for my mistake
Note that $2^8\equiv 1\pmod{51}$, so the given argument breaks down. In fact $2$ as an element of $B$ has order $8$.
Each of our groups has order $32$. So by the Fundamental Theorem, each group is a direct sum of groups of order a power of $2$.
The order of any element of $A$ divides $4$. We will show that the group $B=\left(\mathbb{Z}/51\right)^\times$ has an element of order $16$, so cannot be isomorphic to $A$.
This is true because $B$ is isomorphic to $\left(\mathbb{Z}/51\right)^\times \times\left(\mathbb{Z}/3\right)^\times$, and $\left(\mathbb{Z}/17\right)^\times$ has such an element, since it is cyclic. Alternately, we can find an element of order $16$ explicitly. For instance, $6$ works. But it is enough to exhibit an element of order $8$, and as observed at the beginning, $2$ is such an element.