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**Deveno** An is the subgroup of all even permutations of Sn. to prove/disprove the statement, one usually defines what is meant by the sign or parity of a permutation,

and that sgn:Sn-->({-1,1},*) or parity:Sn--->({0,1},+) is a homomorphism from Sn to a cyclic group of order 2.

in other words, An is/is not normal because (even)(even)(even) = ____ and (odd)(even)(odd) = _____.

the real "meat" of this, is in proving that the sign or parity of a permutation is "well-defined", because a permutation can be written as a product

of transpositions in many, many different ways.

the second problem is a bit easier. for a homomorphism φ:G-->G', ker(φ) is a normal subgroup of G, so |ker(φ)| divides |G|, by lagrange.

by the first isomorphism theorem, φ(G) ≅ G/(ker(φ)). so we have 6 = 20/d, where d is an integer that divides 20.