Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

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.

share|improve this answer
    
Thank you so much for the detail. Now I understand. –  Candy Pelagio Sep 23 '13 at 14:41
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.