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Prove that if $\Omega=\{1,2,3,\cdots \}$ then $S_{\Omega}$ is an infinite group(do not say $\infty !=\infty$).

My attempt : I found the answer here if $\Omega=\{1,2,3,\cdots \}$ then $S_{\Omega}$ is an infinite group, but i have some confusion in this answer , My doubt are given below in red line enter image description here

according to me here $f(m)=m$ match with $f_k's$ that is $f_k(k)=f(k)$. take $k=1$ ,then $f_1(1)=f(1)=1$

why $f(m) =m $ doesn't match with any of the $f_k's ?$

jasmine
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    You seem to be confusing the identity function $f(m)=m$ with the function being defined in the solution, for which $f(m)=m$ for lots of values of $m$ but not the small ones. All they are trying to do is construct an $f$ that is not in the list $f_1,\dots,f_N$ (the logic of the proof shows that this is enough to finish the problem). It's not relevant whether this $f$ satisfes $f(m)=m$ sometimes, or whether other functions wouldn't work. – Greg Martin Aug 16 '20 at 17:51
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    I think, out of the three answers that there are under that MSE question, you have decided to analyse the most complicated and roundabout one. The fact that it was accepted doesn't mean it is necessary the best answer. Look in particular at the least-voted answer: https://math.stackexchange.com/a/77839/700480, it is correct and much simpler. –  Aug 16 '20 at 17:52
  • okss thanks for hints @GregMartin – jasmine Aug 16 '20 at 17:59
  • ya im looking now @StinkingBishop – jasmine Aug 16 '20 at 18:00

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