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I'm trying to understand why how you can determine whether two groups of the form $\mathbb{Z}_n$ are isomorphic to each other. More specifically why is $\mathbb{Z}_2 \oplus \mathbb{Z}_3 \cong \mathbb{Z}_6$ but $\mathbb{Z}_3 \oplus \mathbb{Z}_5 \ncong \mathbb{Z}_{15}$?
Is there some method that I'm missing of determining that two groups are isomorphic?
Point 1. What you are asking is not what you claim to be trying to understand.
Point 2. $\mathbb{Z}_3\oplus\mathbb{Z}_5$ is isomorphic to $\mathbb{Z}_{15}$. What makes you say it isn't?
Point 3. Perhaps you want to compare with $\mathbb{Z}_3\oplus\mathbb{Z}_6$, which is not isomorphic to $\mathbb{Z}_{3\times 6}=\mathbb{Z}_{18}$.
The reason is that while $2$ and $3$ are relatively prime, $3$ and $6$ are not. Every element of $\mathbb{Z}_a\oplus\mathbb{Z}_b$ has order that divides $\mathrm{lcm}(a,b)$. If $\mathrm{lcm}(a,b)\lt ab$, then the group cannot be cyclic, because a finite group of order $k$ is cyclic if and only if there is (at least) one element of order exactly $k$. If every element has order dividing (and hence, less than or equal) to something strictly smaller than the order of the group, then the group cannot be cyclic.
Since $\mathrm{lcm}(a,b)=ab$ if and only if $a$ and $b$ are relatively prime, this tells you that a necessary condition for $\mathbb{Z}_a\oplus\mathbb{Z}_b$ to be cyclic is for $\gcd(a,b)=1$. In order to show that it is enough (sufficient), then notice that the order of $(1,1)$ is exactly $\mathrm{lcm}(a,b)$: $$\begin{align*} k(1,1) = (0,0)\text{ in }\mathbb{Z}_a\oplus\mathbb{Z}_b&\iff (k,k)=(0,0)\text{ in }\mathbb{Z}_a\oplus\mathbb{Z}_b\\ &\iff k\equiv 0\pmod{a}\text{ and }k\equiv 0\pmod{b}\\ &\iff a\text{ divides }k\text{ and }b\text{ divides }k\\ &\iff \mathrm{lcm}(a,b)\text{ divides }k. \end{align*}$$ So the order of $(1,1)$ is exactly $\mathrm{lcm}(a,b)$. Hence, if $\mathrm{lcm}(a,b)=ab$, then $\mathbb{Z}_a\oplus\mathbb{Z}_b$ has (at least) one element of order $ab$, and so it is cyclic.
As a consequence, we have that in (finite) direct sum of cyclic groups, $$\mathbb{Z}_{a_1}\oplus\cdots\mathbb{Z}_{a_k}$$ is cyclic, and isomorphic to $$\mathbb{Z}_{a_1\times\cdots\times a_k}$$ if and only if the $a_i$ are pairwise relatively prime.
Point 4. The question you say you are trying to understand (When two groups of the form $\mathbb{Z}_n$ are isomophic) is: $\mathbb{Z}_n$ is isomorphic to $\mathbb{Z}_m$ if and only if $n=m$. They have to have the same size to be isomorphic; and if they are the same size and both cyclic, then they are isomorphic.
What characterizes the group $\mathbb{Z}_n$ is that
1) It has $n$ elements;
2) it is cyclic.
If you consider $\mathbb{Z}_2\oplus\mathbb{Z}_3$, then you can easily check that $1\oplus 1$ is a generator. Same with $\mathbb{Z}_3\oplus\mathbb{Z}_5$. But if you consider the group $\mathbb{Z}_2\oplus\mathbb{Z}_2$, for example, then every element has order 2, so it cannot be cyclic: it is a group of order 4, but it is not isomorphic to $\mathbb{Z}_4$.
