The problem is to prove or disprove that there is a noncyclic abelian group of order $51$.
I don't think such a group exists. Here is a brief outline of my proof:
Assume for a contradiction that there exists a noncyclic abelian group of order $51$.
We know that every element (except the identity) has order $3$ or $17$. Assume that $|a|=3$ and $|b|=17$. Then I managed to prove that the subgroups generated by $a$ and $b$ only intersect at the identity element, from which we can show that $ab$ is a generator of the whole group, so it is cyclic. Contradiction.
So every element (except the identity) has the same order $p$, where $p$ is either $3$ or $17$.
If $p=17$, take $a$ not equal to the identity, and take $b$ not in the subgroup generated by $a$. Then we can prove that $a^kb^l$ where $k,l$ are integers between $0$ and $16$ inclusive are distinct, hence the group has more than $51$ elements, contradiction.
If $p=3$, take $a$ not equal to the identity and take $b$ not in the subgroup generated by $a$. Then we can prove that $a^kb^l$ where $k,l$ are integers betwen $0$ and $2$ inclusive are distinct. This subgroup has $9$ elements so we can find $c$ that's not of the form $a^kb^l$. Then we can prove that $a^kb^lc^m$ where $k,l,m$ are integers betwen $0$ and $2$ inclusive are distinct. Then this subgroup has $27$ elements so we can find $d$ that's not of the form $a^kb^lc^m$. Then we prove that $a^kb^lc^md^n$ where $k,l,m,n$ are integers between $0$ and $2$ inclusive are distinct, this being $81$ elements. Contradiction.
Best Answer
Using the Sylow theorems would shorten your proof considerably, because from the first Sylow theorem it follows that if a prime divides the order of a group, then the group contains an element of that order. (This eliminates the need to check cases where all elements have order 3 or all have order 17.)
Hence, if $\left|G\right|=51$, then $G$ has an element $a$ of order 3 and an element $b$ of order 17.
Then, as you said, $ab$ has order 51; you can show this directly, without using the proposition for product groups:
Since $\left|G\right|=51$, by Lagrange's theorem, the order of $ab$ divides 51.
Now, since $G$ is abelian, $(ab)^3=a^3b^3=b^3 \neq 1$ because $b$ has order 17, and $(ab)^{17}=a^{17}b^{17}=a^2\neq 1$, because $a$ has order 3.
Hence, $ab$ generates the group, so $G$ is cyclic.