Abstract/Modern AlgebraQuestion:Let G be cyclic and of prime order p. How many automorphisms does G x G have?SolutionAssuming G is group and¿G∨¿p, a primeConsidering Y=G×G is a space vector κ=Ṃp having an action α(x,y) that is (αx,αy); thus for any group homomorphism is given by:m:Y→Y is a κ- linear which lies in GL(Y)=GL2(Ṃp). Therefore, Aut(G)=GL2(Ṃp). Determining the order of the latter is given as follows:g=(ijkl) have a determinant det(g)=il−jkConsidering a two case scenario:1)Assuming i,l≠0 and il=jk that means det(g)=0: for this scenario, there are (p−1)3pairs whichis (i,j,k,l)=(x1,x2,x3,x2x3x1) for x1,x2,x3∈Ṃp×2)If i∨l=0∧j∨K=0; that is il=0∧jk=0 resulting to det(g)=0: for this scenario, there are(2p−1)2 pair. From the two (20 scenario, there are a total of p4 matrices gHence, there are p4−(p−1)3−(2p−1)2Factorizing out p4−(p−1)3−(2p−1)2 gives:(p2−1)(p2−p)Thus, there are:(p2−1)(p2−p)automorphisms
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