I am trying to find two nonisomorphic finite abelian groups with the same order and exponent. I've tried solving this problem for a fews days, but I have had no luck. I tried looking for pairs of such groups of the form $\Bbb{Z}_{mn}$ and $\Bbb{Z}_m \oplus \Bbb{Z}_n$. I've tried searching for examples, but when it comes to numeracy, I am severely lacking. But I fairly convinced that no such pair constitutes an example.
If $m$ and $n$ are relatively prime, then the two groups are actually isomorphic, so this case is relatively uninteresting. Hence, suppose that $m < n$ are not relatively prime, and let $d$ be their GCD. Under this case, there are million subcases and considered for either finding an example or showing no such example exists (e.g., if one could show that $nd < mn$ and $m | nd$, then I believe this would show no such example exists, etc.)
Let $\mathbf{V}=\{(1),(12)(34),(13)(24),(14)(23)\}$ be the four-group, and let $\Gamma_{4}=\langle i\rangle=\{1,i,-1,-i\}$ be the multiplicative cyclic group of fourth roots of unity, where $i^{2}=-1.$ Then these are the two abelian groups with the same order. We will see they are not isomorphic.
If there were an isomorphism $f:\mathbf{V}\rightarrow\Gamma_{4},$ then surjectivity of $f$ would provide some $x\in\mathbf{V}$ with $i=f(x).$ But $x^{2}=(1)$ for all $x\in\mathbf{V},$ so that$$i^{2}=f(x)^{2}=f(x^{2})=f((1))=1,$$contradicting $i^{2}=-1.$ Therefore, $\mathbf{V}$ and $\Gamma_{4}$ are not isomorphic.