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Construct an abelian group of order 12 that is not cyclic.

Can somebody please explain me with examples non-cyclic groups I'm having a hard time understanding.

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You simply need an abelian group of order $12$, with no elements of order $12$. $G=\Bbb Z_6\times\Bbb Z_2$ will do (where $\Bbb Z_n$ denotes the cyclic group of order $n$). As a direct product of cyclic (so abelian) groups, $G$ is again abelian. Given any element $(x,y)\in G$, the order of $(x,y)$ will be the least common multiple of the orders of $x,y.$ The order of $x$ must divide $6$ and the order of $y$ must divide $2,$ so the order of $(x,y)$ is at most $\operatorname{lcm}(6,2)=6.$ But $|G|=|\Bbb Z_6|\cdot|\Bbb Z_2|=6\cdot 2=12.$ Since no element of $G$ has order $12,$ then $G$ is not cyclic.

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  • $\begingroup$ Thank you so much for the detail. Now I understand. $\endgroup$ – Candy Pelagio Sep 23 '13 at 14:41
  • $\begingroup$ Can you give the presentation of this group? $\endgroup$ – MANI SHANKAR PANDEY Jul 5 at 12:01
  • $\begingroup$ @MANI: One would be $$\left\langle a,b\mid a^6,b^2,aba^5b\right\rangle.$$ Is that the sort you're looking for? $\endgroup$ – Cameron Buie Jul 5 at 14:01
  • $\begingroup$ @CameronBuie Yes Sir, I was looking for the same. $\endgroup$ – MANI SHANKAR PANDEY Jul 6 at 4:37
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Consider $\mathbb Z_2 \oplus \mathbb Z_2 \oplus \mathbb Z_3$. It clearly has $2\cdot 2 \cdot 3 = 12$ elements, but every element has order dividing $6$, so there cannot be an element of order $12$, so it isn't cyclic.

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