Why isn't $\mathbb{Z}_2 \times \mathbb{Z}_{30}$ isomorphic to $\mathbb{Z}_{60}$? I know that other groups like $\mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_5$ are isomorphic to $\mathbb{Z}_{60}$. 
 A: Notice that $\mathbb{Z}_{60}$ is cyclic, while $\mathbb{Z}_2\times\mathbb{Z}_{30}$ is not.
That is for example because $\gcd(30,2)\neq 1$.
A: In general, $\mathbb{Z}_n \times \mathbb{Z}_m \cong \mathbb{Z}_{nm} \iff \operatorname{gcd}(n, m) = 1$.  
Even more generally, $\mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \cdots \times \mathbb{Z}_{n_k} \cong \mathbb{Z}_{n_1n_2...n_k} \iff \operatorname{gcd}(n_k, n_j) = 1$ whenever $k \neq j$.

Proof of the first claim:
$\implies$ We will show the contrapositive: suppose $d = \gcd(n, m) \neq 1$.  Then $x = \displaystyle \frac{mn}{d}$ is an integer, and in particular $x = \operatorname{lcm}(m, n)$.  An element $(a, b)$ of a direct product of groups has order $\operatorname{lcm}(|a|, |b|)$.  So consider an element $(a, b) \in \mathbb{Z}_m \times \mathbb{Z}_n$.  The order of $a$ in $\mathbb{Z}_n$ divides $n$, and likewise the order of $b$ divides $m$.  From this we see that the order of $(a, b)$ divides $x$.  Hence, every element of $\mathbb{Z}_m \times \mathbb{Z}_n$ has an order that divides $x$, which is strictly less than the order of the group.   In particular, no element can have order $mn$, so the group cannot be cyclic.
$\ \ $
$\Longleftarrow \ $  Now suppose $\gcd(n, m) = 1$.  The generator $x$ of $\mathbb{Z}_n$ has order $n$, and likewise the generator $y$ of $\mathbb{Z}_m$ has order $m$.  We have $\operatorname{lcm}(|x|, |y|) = |x| \cdot |y| = mn$, which is the order of $\mathbb{Z}_n \times \mathbb{Z}_m$.  Hence, $(x, y)$ generates the group, and it is cyclic. 
$\blacksquare$
The general claim follows per an inductive argument.

Note concerning the above proof: We used the fact that $\displaystyle \operatorname{lcm}(x, y) = \frac{xy}{\gcd(x, y)}$ for all $ x, y \in \mathbb{N}$.
A: There is only one order 2 non-identity $\overline{30}$ in $\mathbb{Z}_{60}$ whereas there are more than one (more precisely, 3) order 2 non-identity elements in $\mathbb{Z}_2\times\mathbb{Z}_{30}$, e.g. $(\overline{1}, 0)$ and $(0, \overline{15})$.
A: $\mathbb{Z}_2\times\mathbb{Z}_{30}\not\cong\mathbb{Z}_{60}$ because, for example, maximum of element's orders in $\mathbb{Z}_2\times\mathbb{Z}_{30}$ not greater than 30, whereas maximum of elements orders in $\mathbb{Z}_{60}$ is 60. Indeed, let $(a,b)\in\mathbb{Z}_2\times\mathbb{Z}_{30}$. Since $a\in\mathbb{Z}_2$, then $2a=0$, hence $30a=15(2a)=0$. Since $b\in\mathbb{Z}_{30}$, then $30b=0$. So, $30(a,b)=(30 a,30 b)=0$, therefore $|(a,b)|\leq 30$. From the other side, in $\mathbb{Z}_{60}$ there exists an element of order $60$. 
This is the argument, mentioned by  @mattbiesecker .
A: Several answers have been provided already. Another one is here: $\mathbb Z_{60}$ contains an element of order 60 but $\mathbb Z_2\times \mathbb Z_{30}$ does not contain any element of order 60. Infact, the maimum possible order of any element in $\mathbb Z_2\times \mathbb Z_{30}$ is lcm$(2,30)=30$.
A: It suffices to look at the subgroups of elements whose order is a power of$~2$. For $\Bbb Z_2 \times \Bbb Z_{30}$ this subgroup is $\Bbb Z_2 \times \Bbb Z_2$ while for $\Bbb Z_{60}$ is it $\Bbb Z_4$, and these subgroups $\Bbb Z_2 \times \Bbb Z_2$ and $\Bbb Z_4$ are not isomorphic.
In general you can decompose by the Chinese remainder theorem $\Bbb Z_{mn}\cong\Bbb Z_m\times\Bbb Z_n$ if (and only if) $m,n$ are relatively prime, so $\Bbb Z_{30}\cong \Bbb Z_2 \times \Bbb Z_{15}$ and $\Bbb Z_{60}\cong \Bbb Z_4\times \Bbb Z_{15}$, which is how the subgroups above are easily found.
A: A funny geometric explanation is that $\text{Spec}(\mathbb Z/2\mathbb Z \times \mathbb Z/30\mathbb Z)$ and $\text{Spec}(\mathbb Z/60\mathbb Z)$ do not have the same number of points. Since $30=2 \times 3 \times 5$, the ring $\mathbb Z/2\mathbb Z \times \mathbb Z/30\mathbb Z$ is a product of $4$ fields hence $\text{Spec}(\mathbb Z/2\mathbb Z \times \mathbb Z/30\mathbb Z)$ has $4$ points. On the other hand, $60=2^2 \times 3 \times 5$, so $\mathbb Z/60\mathbb Z$ is a product of 3 local Artin rings hence it has 3 points. 
