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I believe the the elements of $3 \mathbb{Z}/12 \mathbb{Z}$ are $ \lbrace 0 +12 \mathbb{Z}, 3 +12 \mathbb{Z}, 6 +12 \mathbb{Z}, 9+12 \mathbb{Z} \rbrace = \langle 3 \rangle$, correct? But, what is the factor group isomorphic to and how would I know?

user6259845
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  • What are the possibilities for a group with 4 elements? What is the order of the element 3 + 12Z? I agree with your description of the elements – CJD Dec 08 '17 at 03:31
  • $\Bbb{Z}/12\Bbb{Z}\cong\Bbb{Z}{12}$, and $\Bbb{Z}/3\Bbb{Z}\cong\Bbb{Z}{3}$. Hence $$3\Bbb{Z}/12\Bbb{Z}\cong\Bbb{Z}_{12}/\Bbb{Z}_3$ – QED Dec 08 '17 at 03:33

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Note that $O(3+12\Bbb{Z})=4=O(G)$, where $G=\langle3 \rangle/\langle12 \rangle$ , so $\langle3 \rangle/\langle12 \rangle$ is cyclic and every cyclic group of order $4$ is isomorphic to $\Bbb{Z}_4$

In general, If $k$ divides $n$, then $\langle k \rangle/\langle n \rangle$ is a cyclic group of order $n/k$. So it is isomorphic to $Z_{n/k}$.

Chinnapparaj R
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  • I agree with your answer, but why do you mention the order of 6 + 12Z? Is it true that if the orders of elements in two groups are the same, then the groups are isomorphic? I don't think that's true. – CJD Dec 08 '17 at 03:40
  • ok! I understand .....I have edit my answer. Is it correct now? – Chinnapparaj R Dec 08 '17 at 03:51
  • Yeah, that looks good! I think it would be most natural to say that the group has 4 elements, and has an element of order 4, hence it's cyclic. (So you also don't have to mention the order of 9 + 12Z.) I looked up the question I asked about orders and the smallest counter-example has 16 elements: https://math.stackexchange.com/questions/1296833/if-i-know-the-order-of-every-element-in-a-group-do-i-know-the-group – CJD Dec 08 '17 at 03:56