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The cyclic group $\mathbb Z_6$ has a subgroup of order $3$ so I can deduce that $\mathbb Z_3$ is a subgroup of $\mathbb Z_6$. On one hand, it seems false to me cause the group operations of the above groups are not the same. On the other hand, the subgroup of order $3$ is isomorphic to $\mathbb Z_3$. I am confused...

José Carlos Santos
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Hani
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3 Answers3

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No, you cannot deduce that $\Bbb Z_3$ is a subgroup of $\Bbb Z_6$. What you can deduce is that $\Bbb Z_6$ has a subgroup which is isomorphic to $\Bbb Z_3$.

José Carlos Santos
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  • I am going deep in elementary abstract algebra. I read this https://math.stackexchange.com/a/105440/809841. it made me to ask myself the above question. Z6 has a subset of order 3 so this subset can be Z3. Am I right? Thx – Hani Jul 19 '20 at 06:43
  • When you wrote subset did you mean subgroup? – José Carlos Santos Jul 19 '20 at 06:45
  • Sorry...Yes I meant subgroup. In fact, the cited link above is confusing me about the operations. If it is right (of course it is) then a set wit 6 elements can be considered as Z6 and every subset of order 5 of it can be considered as Z5 and a contradiction! I feel I am missing a prominent point. Thx – Hani Jul 19 '20 at 06:49
  • What that answer is saying is not that a set with $6$ elements can be considered $\Bbb Z_6$. It says that on any set with $6$ elements you can define a group structure on it that it will make it isomorphic to $\Bbb Z_6$. – José Carlos Santos Jul 19 '20 at 06:52
  • Let me reflect... – Hani Jul 19 '20 at 06:57
  • Thx a lot for the time! and sorry for asking such this simple question... – Hani Jul 19 '20 at 12:53
  • Apparently, I can not put any comments on Arturo's neat answer so, may I ask something relevant to my question? It is between your answer above and his. Thx – Hani Jul 19 '20 at 12:58
  • Sure . The worst that can happen is that I suggest that you post that question as a new question. – José Carlos Santos Jul 19 '20 at 13:23
  • I asked as you suggested – Hani Jul 19 '20 at 16:06
  • And I have provided an answer to that question. – José Carlos Santos Jul 19 '20 at 16:18
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The set $\mathbb Z/3\mathbb Z$ is not even a subset of $\mathbb Z/6\mathbb Z$.

There are three members of $\mathbb Z/3\mathbb Z$. Explicitly, they are:

\begin{align} \{\ldots,{-7},{-4},{-1},2,5,\ldots\}&\in\mathbb Z/3\mathbb Z\\ \{\ldots,{-6},{-3},0,3,6,\ldots\}&\in\mathbb Z/3\mathbb Z\\ \{\ldots,{-5},{-2},1,4,7,\ldots\}&\in\mathbb Z/3\mathbb Z \end{align}

There are six members of $\mathbb Z/6\mathbb Z$:

\begin{align} \{\ldots,{-14},{-8},{-2},4,10,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-13},{-7},{-1},5,11,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-12},{-6},{0},6,12,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-11},{-5},{1},7,13,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-10},{-4},{2},8,14,\ldots\}&\in\mathbb Z/6\mathbb Z\\ \{\ldots,{-9},{-3},{3},9,15,\ldots\}&\in\mathbb Z/6\mathbb Z \end{align}

Note that there is no $x$ such that $x\in\mathbb Z/3\mathbb Z$ and $x\in\mathbb Z/6\mathbb Z$. The two sets have no members in common.

Since it isn't a subset, $\mathbb Z/3\mathbb Z$ is not a subgroup of $\mathbb Z/6\mathbb Z$ either.

Chris Culter
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$\Bbb{Z}_6$ has a subgroup $\{[0],[2],[4]\}$ which is isomorphic to $\Bbb{Z}_3$, but $\Bbb{Z}_3$ is not even a subset of $\Bbb{Z}_6$.