How to prove that $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_{15}$ is not cyclic. Generally, to prove that a group is not cyclic I check if there exists an element in the group that is of the same order as the group.  If there is not such an element than the group is not cyclic as it does not have a generator.  However, the group $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_{15}$, where $\times$ is the Cartesian product, has 60 elements in it.  Ain't nobody got time to check the order of 60 elements.  Is there a clever way to determine if this group is cyclic?
Also, if there exists an element in a group of the same order as the group then is that element automatically the generator, or would I have to check whether that element actually generates the group? 
 A: A cyclic group of order $n$ has a unique subgroup of order $d$ for every divisor of $n$. Your group has order $15\times 4$ and has two distinct subgroups of order $2$ (in fact three but the point here is it has more that one). 
A: It is easy, I believe, to check that for any element
$$(a,b,c)\in\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_{15}\;,\;\;30\cdot(a,b,c)=(0,0,0)$$
Other way:
$$\Bbb Z_2\times\Bbb Z_2\times\Bbb Z_{15}\cong\Bbb Z_2\times\Bbb Z_{30}$$
and $\;\Bbb Z_n\times\Bbb Z_m\;$ is never cyclic if $\;gcd(n,m)>1\;$ .
A: Simply: there is no element of order $4$, hence such a group with $60$ elements cannot be cyclic.
A: For direct products to be cyclic, both groups must be cyclic, and $Z_2$ x $Z_2$ is not cyclic.
A: It's actually enough to prove that the group $\mathbb Z_2\times \mathbb Z_2$ isn't cyclic, since if $(a, b, c)$ generated $\mathbb Z_2\times \mathbb Z_2\times\mathbb Z_{15}$, then $(a, b)$ would generate $\mathbb Z_2\times\mathbb Z_2$, or since, more generally, every subgroup of a cyclic group is cyclic.
