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?

• If the order of an element $a$ is equal to $|G|$, then you can conclude $G=<a>$ – Peter Nov 5 '16 at 21:58
• @Alephnull Not true in general, for example $\mathbb Z_3\times\mathbb Z_5$, which is isomorphic to $\mathbb Z_{15}$ , is cyclic. – Peter Nov 5 '16 at 22:26
• @Alephnull Linear algebra? How come? What this "isn't dimension 1"?? – DonAntonio Nov 5 '16 at 22:31
• Subgroups of cyclic group are cyclic. And it is easy to see that $Z_2\times Z_2$ is not cyclic. So, our group is not cyclic. – mesel Nov 5 '16 at 22:46
• @Alephnull Could you please clear up the questions I asked you in my past comment? – DonAntonio Nov 5 '16 at 22:55

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).

• This is very good to save time, at least in this case...and that is an if and only if claim: a finite abelian group is cyclic iff it has one unique subgroup of any order dividing the group's. – DonAntonio Nov 5 '16 at 22:03
• Would two subgroups of order 2 be the cyclic groups generated by (0,1,0) and (1,0,0)? And since these subgroups are not the same then the group is not cyclic? – PiccolMan Nov 5 '16 at 22:52
• @PiccolMan Exactly. – Pedro Tamaroff Nov 5 '16 at 22:53

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 $\;gcd(n,m)>1\;$ .

• Another way it so suppose $(a,b,c)$ generates $\Bbb Z_2\times \Bbb Z_2\times \Bbb Z_{15}$, see that $a=b=1$ and thus conclude that it is not cyclic since we reach a contradiction. – Astyx Nov 5 '16 at 21:59
• Thanks for responding. I chose the other response as the best answer because it was simpler, but thank you. – PiccolMan Nov 5 '16 at 22:58

Simply: there is no element of order $4$, hence such a group with $60$ elements cannot be cyclic.

For direct products to be cyclic, both groups must be cyclic, and $Z_2$ x $Z_2$ is not cyclic.

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.