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I am reading a first course in algebra and there is an example saying that "find all the subgroups of $\Bbb{Z}_2\times\Bbb{Z}_6$ and decide which of them are cyclic. I know that $$\Bbb{Z}_2\times\Bbb{Z}_6=\{(0,0),(0,1),\ldots,(0,5),(1,0),\ldots,(1,5)\}$$ consisting of $12$ elements and since its order is $12$, the subgroups can be of order $1$, $2$, $3$, $4$, $6$ or $12$.

But I do not know how to find the subgroups and decide whether which of them are cyclic. Can anyone help?


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Could it be easier for you if you write $\,\Bbb Z_2\times\Bbb Z_2\times \Bbb Z_3\,$ ? – DonAntonio Mar 24 '13 at 21:45
up vote 12 down vote accepted

Hint: Recall the theorem highlighted below, and note that it follows that $$\quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 6} \quad \cong \quad \mathbb Z_{\large 2} \times \mathbb Z_{\large 2}\times \mathbb Z_{\large 3}$$

This might help to make your task a bit more clear, noting that each of $\mathbb Z_2, \; \mathbb Z_3,$ and $\,\mathbb Z_6 \cong \mathbb Z_2 \times \mathbb Z_3$ are cyclic, but $\;\mathbb Z_2 \times \mathbb Z_2,\;$ of order $\,4,\,$ is not cyclic. Indeed, there is one and only one group of order $4$, isomorphic to $\mathbb Z_2\times \mathbb Z_2$, i.e., the Klein $4$-group.

Theorem: $\;\mathbb Z_{\large mn}\;$ is cyclic and $$\mathbb Z_{\large mn} \cong \mathbb Z_{\large m} \times \mathbb Z_{\large n}$$

if and only if $\;\;\gcd(m, n) = 1.$

This is how we know that $\mathbb Z_6 = \mathbb Z_{2\times 3} \cong \mathbb Z_2\times \mathbb Z_3$ is cyclic, since $\gcd(2, 3) = 1.\;$

It's also why $\,\mathbb Z_2\times \mathbb Z_2 \not\cong \mathbb Z_4,\;$ and hence, is not cyclic, since $\gcd(2, 2) = 2 \neq 1$.

Good-to-know Corollary/Generalization:

The direct product $\;\displaystyle \prod_{i = 1}^n \mathbb Z_{\large m_i}\;$ is cyclic and $$\prod_{i = 1}^n \mathbb Z_{\large m_i}\quad \cong\quad \mathbb Z_{\large m_1m_2\ldots m_n}$$ if and only if the integers $m_i\,$ for $\,1 \leq i \leq n\,$ are pairwise relatively prime, that is, if and only if for any two $\,m_i, m_j,\;i\neq j,\;\gcd(m_i, m_j)=1$.

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Thanks, I see that Z2,Z3 and Z6 are cyclic subgroups, but how to find whether there are other subgroups? – bigO Mar 24 '13 at 21:53
Can $\mathbb Z_2 \times \mathbb Z_2$ of order 4 possibly be cyclic? See the "Recall...if and only if..." – amWhy Mar 24 '13 at 21:56
There're lots more: if $\,a,b\,$ are the generators of $\,\Bbb Z_2\,,\,\Bbb Z_6\,$ , then $\,\langle\,(a,b^2)\,\rangle\,$ is a group of order $\,3\,$ , $\,\langle\,(a,b^3)\,\rangle\,$ is a group of order $\,2\,,$ etc. – DonAntonio Mar 24 '13 at 21:56
Oh, sorry @amWhy: my last comment was directed to bigO, not to you. – DonAntonio Mar 24 '13 at 21:59
@amWhy: This certainly deserves a nice answer merit badge! :-) +1 – Amzoti Apr 18 '13 at 1:05


  1. To find the cyclic subgroups, calculate $\langle g \rangle$ for each $g \in \mathbb{Z}_2 \times \mathbb{Z}_6$.

  2. Every group of prime order is cyclic.

  3. Every abelian group of order $6$ is cyclic. (Alternatively, use the fact that every element of a subgroup of order $6$ must have order $1$, $2$ or $3$).

  4. The group $\mathbb{Z}_2 \times \mathbb{Z}_6$ has exactly one subgroup of order $4$. (Use the fact that any element of the subgroup must have order $1$, $2$ or $4$).

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Look at some elements and see what subgroups they generate alone.

E.g. $(0,1)$ generates the $\{0\}\times\Bbb Z_6$ copy, $(0,2)$ generates a $\Bbb Z_3$ (just as $2$ in $\Bbb Z_6$), $(1,3)$ generates a $\Bbb Z_2$.

Then you can look for subgroups generated by $2$ elements, e.g. $(0,2)$ and $(1,0)$ together generate $\Bbb Z_2\times\Bbb Z_3$, and so on...

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