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In proving this Sylow's Theorem:

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The proof starts by assuming inductively the existence of Sylow's p-subgroups as following:

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Isn't it circular?

J. W. Tanner
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Ari Royce Hidayat
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    did you catch that the proof uses induction on the order of the group? – J. W. Tanner Dec 30 '19 at 03:53
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    No because they prove if $p^{\alpha - 1}$ holds then it holds for $p^{\alpha}$. – CyclotomicField Dec 30 '19 at 03:54
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    When you see something like "proceed by induction on $n$", it means that we first assume the statement $P(k)$ to be proven holds true for all $k\leq n$, and proceed to show that $P(n+1)$ is true. Here induction on $|G|$ means induction on the order of the group, so we are assuming the theorem holds for all groups with order less than $|G|$ for a particular group $G$, then showing it must work for $G$ too. There's no circularity here. – YiFan Tey Dec 30 '19 at 04:02

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This is the usual structure of a strong induction proof. The intuitive idea is that if the statement fails for some set of numbers, there is a smallest one that it fails for. If we show that (it is true for $n=1$ and) for all $n$, the fact that it is true for all numbers $1$ through $n$ implies it is true for $n+1$ shows there is no smallest number that it fails for, so it must be true for all $n$. Here the induction is on the size of $G$. The claim is that if there is a group without a Sylow subgroup, there is one of minimum order. When we consider that group, all smaller groups have Sylow subgroups. We derive a contradiction and conclude that there is no smallest group without a Sylow subgroup.

Ross Millikan
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  • But how could we be sure that if it started from some true assumption, then if some condition is not-true, that not-true would finally be revealed? – Ari Royce Hidayat Jan 04 '20 at 09:20
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    That possibility is ruled out by the proof. If the smallest group without a Sylow subgroup were of order $k$, all groups of size $1$ through $k-1$ would have Sylow subgroups. There would have to be something wrong in the proof to allow one of size $k$. – Ross Millikan Jan 04 '20 at 14:48
  • But if we have $|G| = pm$, and the smallest one without a Sylow subgroup is $G$ or ${1}$, how could the proof prove it wrong? – Ari Royce Hidayat Jan 05 '20 at 11:20
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    Without the earlier text I cannot fully follow the proof. One paragraph shows that $p$ cannot divide the order of the center and the next paragraph shows that $p$ must divide the order of the center. That contradiction shows that a group of size $pm$ s not the smallest group without a Sylow subgroup. I recommend https://math.stackexchange.com/questions/19485/dominoes-and-induction-or-how-does-induction-work/19488#19488 for a nice outline of induction proofs. – Ross Millikan Jan 05 '20 at 14:37
  • It's just there is no earlier text :D Anw thanks a lot, from the answer you refer, now I think I know what I missed. It has been mentioned in the text that if $|G| = 1$ then there is nothing to prove, which is the base case of the induction used. (I missed that a group of order $p^0 1$ is also a Sylow subgroup.) – Ari Royce Hidayat Jan 05 '20 at 17:29
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Note that it first assumes the existence of Sylow subgroups for all groups of order less then $|G|$. By that assumption, it proves the existence of Sylow subgroup of $G$. Since $|G| = 1$ is a group of order less than any other groups, and $G$ is definitely a Sylow subgroup, it completes the proof.

Ari Royce Hidayat
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