October 11th
Today I learned that the commutator subgroup (subgroup generated by $aba^{-1}b^{-1}$) of $S_n$ is $A_n,$ for $n\ge3.$ Let the commutator subgroup be $C=\langle \sigma\tau\sigma^{-1}\tau^{-1}:\sigma,\tau\in S_n\rangle,$ and we'll show $C=A_n.$
In one direction, $C\subseteq A_n$ is not hard. Simply note that for any $\sigma\tau\sigma^{-1}\tau^{-1}\in C,$ we have\[\op{sgn}\left(\sigma\tau\sigma^{-1}\tau^{-1}\right)=\op{sgn}(\sigma)\op{sgn}(\tau)\op{sgn}{\sigma}^{-1}\op{sgn}(\tau)^{-1}=1\]because $\op{sgn}$ is a homomorphism. From this we see $\sigma\tau\sigma^{-1}\tau^{-1}\in A_n,$ so $C\subseteq A_n.$ It remains to show $A_n\subseteq C.$
As a lemma, we show that $A_n$ is generated by $3$-cycles; for clarity, this is why $n\ge3$ matters. Indeed, $A_n$ is comprised of all even permutations, so it is generated by permutations made up of two transpositions, say $(a,b)(c,d).$ We do casework on the overlap.
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If $(a,b)=(c,d),$ then this is the identity, which is generated by $3$-cycles.
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If only one letter in each is the same, say $b=c$ with $a\ne d,$ then we see \[(a,b)(b,d)=(a,b,d)\] is already a $3$-cycle.
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If all letters are different, then we write \[(a,b)(c,d)=(a,b)\circ(b,c)(b,c)\circ(c,d)=[(a,b)(b,c)]\circ[(b,c)(c,d)],\] which reduces to the previous case.
It follows that the generators of $A_n$ can be expressed as products of $3$-cycles, so $A_n$ can be expressed as products of $3$-cycles.
With the lemma in mind, it suffices to express $3$-cycles as a commutator, which will force $A_n\subseteq C$ because we just showed $A_n$ is generated by those $3$-cycles. And indeed,\[(a,b,c)=(a,c)(b,c)(a,c)(b,c)=(a,c)(b,c)(a,c)^{-1}(b,c)^{-1}\in C.\]This completes the proof that $A_n\subseteq C,$ and so we are done.
In other news, happy birthday to Adison Lampert!