As we all know free groups are always infinite, as they have no restrictions on their base sets. I have confusion as $S_3$ and $\mathbb{Z}_6$ both are groups of order $6$. What can we say about the free groups generated by these groups?

Are the isomorphic?

If they are not isomorphic, then what is the reason behind all that?

I have studied so many questions on this site about free groups but didn't get any question like this. Someone told me not to ask such question here, but I am asking because this is my level of algebra. I can't understand that,s why I am asking. I am studying Combinatorial group theory by Wilhelm Magnus, Abraham Karrass and Donald Solitar and I came across such confusion. Please help me out to short out this problem.

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    $\begingroup$ This question needs some context (why are you asking it? Is it written down somewhere?). As it stands, a group does not generate a free group, but rather a set does. Possibly you meant to ask something different, I don't know. $\endgroup$
    – user1729
    Sep 4, 2019 at 10:59
  • $\begingroup$ actually, I am studying a new structure(multiplicative Lie algebra) born by Ellis. He told that if $G$ is an abelian group then we can generate a free multiplicative Lie algebra $L(G_{ab})$ such that the underlying group is abelian. I was confused why this is happening. But now its ok. $\endgroup$ Sep 4, 2019 at 11:24
  • $\begingroup$ The free group generated by $\emptyset$ is not infinite, but also is not an all too exiting group. Makes sense to exclude that one. $\endgroup$
    – Con
    Sep 4, 2019 at 12:02
  • $\begingroup$ @Priya I don't understand. I think I can guess what you are after, but it would be helpful if you gave the definition/issue that you are trying to understand. $\endgroup$
    – user1729
    Sep 4, 2019 at 16:38
  • $\begingroup$ @user1729 dear sir, this is the reference paper at which I am working. In Paper at page $5$ Proposition $2$ they are saying that a free multiplicative Lie algebra is similar to usual free Lie algebra on an abelianized group. I am not able to understand this phenamenon, thats why I asked it here. $\endgroup$ Sep 5, 2019 at 14:20

1 Answer 1


Yes they are isomorphic. When we say that $S$ is a generating set of a free group we just mean that it is a set of symbols which have inverses.

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    $\begingroup$ @MANI SHANKAR PANDEY But free groups are generated by sets. What does generated by $\mathbb{Z}_n$ mean? $\endgroup$
    – 1123581321
    Sep 4, 2019 at 10:48
  • $\begingroup$ thanks for your kind answer. But then how we can say that both $S_3$ and $\mathbb{Z}_6$ are quotients of the same free group? $\endgroup$ Sep 4, 2019 at 10:51
  • $\begingroup$ @Priya Pandey So, let $X$ be a set of $6$ elements and the free group $F=F(X)$ generated by $X$. Then $S_3\cong\dfrac{F}{<<R>>}$ and $\mathbb{Z}_6\cong\dfrac{F}{<<Q>>}$. Can you find some $R$ and some $Q$? $\endgroup$
    – 1123581321
    Sep 4, 2019 at 10:54
  • $\begingroup$ Yes, now I got the key. again thanks for your answer. $\endgroup$ Sep 4, 2019 at 10:55
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    $\begingroup$ @giannispapav Sorry, I was little confused, now it is ok. $\endgroup$
    – MANI
    Sep 4, 2019 at 11:01

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