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Here is the problem, I thought it the I've been thinking for a long time, but I still don't have any ideas.

Problem: Let A(S) be the group of all bijective functions on S with composition as its binary operation. then A(S) is finitely generated if and only if S is finite set.

I know that proving from the right to the left is relatively easier, but when it comes to proving from the left to the right, I really don't have any ideas. I hope someone can help.

Shaun
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Miicky
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Assuming that $S$ is infinite, $A(S)$ is uncountable, but if it were finitely generated, it would be countable. So, if $A(S)$ is finitely generated, $S$ must be finite. (Added later: Going in the other direction, if $S$ is finite, $A(S)$ is finite, so it's finitely generated.)

David Moews
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  • Oh! nice answer! Thx a lot!!!! Can I ask how do you prove right to left? – Miicky Sep 23 '23 at 17:28
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    If $S$ is finite, $A(S)$ is finite, so it's finitely generated. – David Moews Sep 23 '23 at 17:30
  • @Miicky In fact, if $S$ is finite, then $A(S)$ is $2$-generated. Namely, if $S$ has $n$ elements ($n\ge3$), then $A(S)$ is generated by an $n$-cycle and a wisely chosen transposition. – bof Sep 24 '23 at 04:01