By Nicholas Jackson

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X n ) = ∏ ( xi − x j ) 1 i< j n = ( x1 − x2 )( x1 − x3 ) . . ( x1 − xn )( x2 − x3 ) . . ( xn−1 − xn ). Given a permutation σ ∈ Sn , we define σ( P)( x1 , . . , xn ) = P( xσ(1) , . . , xσ(n) ). The permuted polynomial σ ( P)( x1 , . . , xn ) has exactly the same factors ( xi − x j ) as P( x1 , . . , xn ), except that some of the variables xi have been permuted, and some of the factors will change sign. We may therefore define the sign of σ to be the quotient σ ( P)( x1 , . . , xn ) = ±1.

8 Write out the group multiplication table for Z2 ⊕ Z5 . 9 Let G be a group. Show that if ( ab)2 = a2 b2 for all a, b ∈ G, then G must be abelian. 10 Let G be a nonempty set and let ∗ : G × G → G be an associative binary operation. Suppose that there is a distinguished element e ∈ G such that: (a) e ∗ x = x for all x ∈ G. ) (b) For any x ∈ G, there exists y ∈ G such that y ∗ x = e. ) Then show that G = ( G, ∗) is a group. 11 Let G be a group, and suppose that g ∈ G and m, n ∈ Z. Prove that ( gm )n = gmn and gm gn = gm+n .

Lk ). That is, the order of π is the lowest common multiple of the lengths of the disjoint cycles σ1 , . . , σk . Proof First, observe that π n = (σ1 . . σk )n = σ1n . . σkn since disjoint cycles commute. Then if |π | = n this means that π n = ι and hence σ1n . . σkn = ι. Now suppose that σi = ( x1 . . xli ). Then π n = ι implies that π n ( x j ) = x j for 1 j li . So (σ1n . . σkn )( x j ) = x j , and since the σs are disjoint cycles, we know that no other cycle apart from σi affects x j and therefore σi ( x j ) = x j for 1 j li .