Let $G$ be an abelian group and $\phi: S_n\mapsto G$ a homomorphism. Then $A_n\subset\ker\phi$. Let $G$ be an abelian group and $\phi: S_n\mapsto G$ a homomorphism. Then $A_n\subset\ker\phi$.
Hint: Compute $(1a)(1b)(1a)(1b)$ for $a\neq 1\neq b$ and show that every element in $A_n$ can be written as a product of the elements in the form $\sigma\tau\sigma^{-1}\tau^{-1}$ with $\sigma,\tau\in S_n$.
I get that the alternating group is a normal subgroup of the symmetric group. I don't understand what the author means by the first part of the hint, namely calculating $(1a)(1b)(1a)(1b)$ as in I don't get the notation, what to $a,b$ belong to? Probably something obvious I am missing. Any further info would also be nice but mainly the notation.