# Prove Alternating group is in kernel of homomorphism from symmetric to abelian group.

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.

• Yeah I know what transpositions are, was not sure what a,b where supposed to be anyway I get it now, thanks. So from what I see (1a)(1b)(1a)(1b)=(1ab) ? – Sam Papadoulos Dec 4 '16 at 20:47
• Yes, looks good. – Dietrich Burde Dec 4 '16 at 20:55
• So using gyazo.com/1a38a1a6bb40ea42f907b58a5083f4a8 we have that all element of An can be expressed as a product of 3 cycles but for any 3 cycle abc we have abc= (ab)(ac)(ab)(ac) but ab=(ab)^-1 and likewise for ac so when taking the homomorphism φ of some x in An φ(χ)=φ(αβ)*φ(ac)*φ(ab)*φ(ac)...=φ(αβ)*φ(αβ)*φ(ac)*φ(ac)..=1 (because abelian). Is this correct? – Sam Papadoulos Dec 4 '16 at 23:05