In another thread I asked for the proof of the following proposition:

Proposition: Let $G_1,\dots,G_n$ be finite groups. Then $G_1\times\dots\times G_n$ is cyclic iff $G_1,\dots,G_n$ are cyclic and $\gcd(\operatorname{ord} G_i,\operatorname{ord} G_j)=1$ if $i\neq j$.

My problem lies with the counterintuitive formula of finding if the direct product of two groups is indeed cyclic. $\gcd(\operatorname{ord} G_i,\operatorname{ord} G_j)=1$ if $i\neq j$.

In an example on the material I am studying it follows:

Case I

$\mathbb{Z}_4\times\mathbb{Z}_3\times \mathbb{Z}_{10}$ is not cyclic.

I know that $4,3,10$ have at least common multiple on this case $60$. I know if the group is cyclic the order of $\mathbb{Z}_4\times\mathbb{Z}_3\times\mathbb{Z}_{10}$ is $60$.


Why is $\mathbb{Z}_4\times\mathbb{Z}_3\times\mathbb{Z}_{10}$ is not cyclic?

Case II

There is also the case: $\mathbb{Z}_2\times\mathbb{Z}_6$, generated by $ord(1_4,1_6)=12$ The $gcd(4,6)=2\neq 1$


Is $\mathbb{Z}_2\times\mathbb{Z}_6$ not cyclic?


What is wrong with my interpretation of the theorem statement?

Thanks in advance!

  • $\begingroup$ $LCM(4,3,10)=60$ $\endgroup$ – idok Dec 22 '17 at 15:53

The group $\mathbb{Z}_4\times\mathbb{Z}_3\times\mathbb{Z}_{10}$ has $120$ elements, but every element has order a divisor of $60$, so no element can be a generator.

Similarly for $\mathbb{Z}_2\times\mathbb{Z}_6$, which has $12$ elements, but maximal order of elements $6$.

If $g\in G$, then the subgroup generated by $g$ has the same number of elements as the order of $g$.


Actually, $\;\mathbf Z/4\mathbf Z\times\mathbf Z/3\mathbf Z\times\mathbf Z/10\mathbf Z\simeq\mathbf Z/12\mathbf Z\times\mathbf Z/10\mathbf Z$

Now this group has order $120$, and if it were cyclic, it would contain an element of order $120$. However the order of an element $(a\bmod 12,b\bmod 10)$ is $\,\operatorname{lcm}(a\bmod 12,b\bmod 10)$, which is a divisor of $\,\operatorname{lcm}(12,10)=60$.

Second question:

No, $\mathbf Z/2\mathbf Z\times\mathbf Z/6\mathbf Z$, is not cyclic for he same reason: it should contain an element of order $12$. You can easily check each of the $12$ elements have order a divisor of $6$.


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