Noncyclic Abelian Group of order 51

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.

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.

You're right.

If you know the theorem that classifies finite abelian groups, then the only possible abelian groups of order $51$ are $\mathbb Z/51 \mathbb Z$ which is cyclic and $\mathbb Z/3 \mathbb Z \times \mathbb Z/17 \mathbb Z$ which is also cyclic because $3$ and $17$ are prime and $\gcd(3,17)=1$ so $\mathbb Z/3 \mathbb Z \times \mathbb Z/17 \mathbb Z \simeq \mathbb Z/51 \mathbb Z$.

Any finite abelian group is isomorphic to a group of the form $$\Bbb Z_{p_1^{a_1}}\times \Bbb Z_{p_2^{a_2}}\times\cdots\times \Bbb Z_{p_n^{a_n}}$$ where $p_i$ are (not necessarily distinct) primes. The order of such a group is $p_1^{a_1}\cdots p_n^{a_n}$. How many ways can this be done for $51$?