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it seems pretty trivial, but I have trouble of showing it.

I also wonder of good approach of proving that a group is not finitely generated.

d_e
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    please explain why is the down-vote and 2 close votes. I'm asking this question for self-learning. I did try to solve it but without much of success. my tries are only last for few seconds till I understood they are wrong. I would be happy to receive hints. as I have no direction now. – d_e Jul 28 '15 at 10:57
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    The cardinality is wrong. That is: the symmetric group has the cardinality of the power set, but were it finitely generated it would be countable. (Note: I did not vote the question down. The group you are looking at is very hard to get a handle on and I can certainly accept that you hit a stone wall). – lulu Jul 28 '15 at 11:02
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    As a curiosity: The subgroup G consisting of all permutations which only move finitely many elements is obviously not finitely generated (as any list of generators would fix infinitely many elements), but this does not solve your problem as, sadly, finitely generated groups can have infinitely generated subgroups. – lulu Jul 28 '15 at 11:05
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    @lulu, thanks , the issue of cardinality is indeed worth consideration.it is indeed seem to solve the issue. so much thanks. – d_e Jul 28 '15 at 11:14
  • @lulu: I think this is a clear case of comment --> answer :-). (I also didn't downvote. Returning after a ~ 2-year absence, I find that rapid downvoting of questions that don't show much effort has become a lot more aggressive than it used to be. This can be appropriate in some cases of routine calculations thrown at us in copy-paste style, but I find it annoying that it seems to be done almost automatically, irrespective of the difficulty of the question or of remarks like wondering about a good approach.) – joriki Jul 28 '15 at 11:21
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    @joriki I agree completely. There's a world of difference between someone posting their related rates homework and someone who is drawing a blank in the face of a hard to grasp idea. I'll post my comment as an answer now. – lulu Jul 28 '15 at 11:24

1 Answers1

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The cardinality is off. The symmetric group has the cardinality of the power set, see, for example, https://mathoverflow.net/questions/27785/cardinality-of-the-permutations-of-an-infinite-set. But a finitely generated group is countable.

As noted in the comments: it is interesting to remark that the subgroup consisting of all permutations which move only finitely many elements is clearly not finitely generated (as any list of generators would jointly fix infinitely many elements). This, however, does not solve the problem as it is perfectly possible for a finitely generated group to have an infinitely generated subgroup (famously, the free group on 2 letters contains an infinitely generated subgroup).

Community
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lulu
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  • Hi lulu, I now returns to read that answer, however, I find no example of a subgroup of the free group on 2 letters , that is infinitely generated. – d_e Jan 10 '16 at 13:36
  • http://math.stackexchange.com/questions/983480/commutator-subgroup-of-rank-2-free-group-is-not-finitely-generated – lulu Jan 10 '16 at 13:44