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- 11. Show that defined by is not a homomorphism.6. Prove that if is a permutation on , then is a permutation on .Let G=1,i,1,i under multiplication, and let G=4=[ 0 ],[ 1 ],[ 2 ],[ 3 ] under addition. Find an isomorphism from G to G that is different from the one given in Example 5 of this section. Example 5 Consider G=1,i,1,i under multiplication and G=4=[ 0 ],[ 1 ],[ 2 ],[ 3 ] under addition. In order to define a mapping :G4 that is an isomorphism, one requirement is that must map the identity element 1 of G to the identity element [ 0 ] of 4 (part a of Theorem 3.30). Thus (1)=[ 0 ]. Another requirement is that inverses must map onto inverses (part b of Theorem 3.30). That is, if we take (i)=[ 1 ] then (i1)=((i))1=[ 1 ] Or (i)=[ 3 ] The remaining elements 1 in G and [ 2 ] in 4 are their own inverses, so we take (1)=[ 2 ]. Thus the mapping :G4 defined by (1)=[ 0 ], (i)=[ 1 ], (1)=[ 2 ], (i)=[ 3 ]