Here are the questions Let h : G rightarrow G\' be a group homomorphism, with G

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Let h : G rightarrow G' be a group homomorphism, with G and G' groups. Assume G is cyclic and h is surjective. Prove that G' is cyclic. Hint. Say G = [g] for some (fixed) g epsilon G. Let x epsilon G' be arbitrary. Use the assumption that h is onto to realize x as a power of a fixed element (which one?) in G'. Show your work. Let h : G -> G' be a homomorphism (with G, G' groups) and G' abelian. Denote the identities of G and G' by e and e' respectively. Let N = {x epsilon G | G (h(x))2 = e'}. Prove N

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4. h:G->G' onto homomorphism. As given in hint since G is cyclic suppose G = , i.e group G is generated by element g. Look at g' = h(g) in G'. Now we shall show that g' generated G' and hence that G' is cyclic. Let x be any element of G' Then there is an element y in G such that h(y) = x (since g is onto, note that we are not concerned with uniqueness of this element, we just need one such element) Now G is generated by g. So y = g^n for some n>=0, integer. Now x = h(y) = h(g^n) = [h(g)]^n (since h is a homomorphism) = (g')^n This shows that any element x in G' is equal to (g')^n for some integer n>=0. Thus G' = as claimed and thereby G' is cyclic. 5. (1) Let x and y be elements of N. Then (h(x))^2 = e' = (h(y))^2 => (h(xy))^2 = h(xy).h(xy) = h(x).h(y).h(x).h(y) = h(x).h(x).h(y).h(y) (h(x), h(y) being elements of G' commute since G' is abelian) = (h(x))^2.(h(y))^2 = e'.e' = e' Thus xy is in N whenever x and y are in. Now if x is an element of N. (h(x^-1))^2 = (h(x)^-1)^2 = ((h(x))^2)^-1 = (e')^-1 = e'. This shows that x^-1 is in N whenever x is in N. Thus N is closed with respect to the group operation as well as with respect to inverses. Hence N is a subgroup of G. (2) Now let x be any element of G and y is any element of N Then (h(yxy^-1))^2 = (h(y).h(x).h(y^-1))^2 = h(y).h(x).h(y^-1). h(y).h(x).h(y^-1) = [h(y).h(y^-1)].[h(y).h(y^-1)].[h(x).h(x)] (G; is abelian so we can commute elements and write in any order) = e'.e'.e' (since h(y).h(y^-1) = h(y).h(y)^-1 (h homomorphism) = e', and (h(x))^2 = e') = e'. Thus yxy^-1 is in N for x in N, y in G. This shows yNy^-1 is contained in N for all y in G. Thus {yNy^-1 : y is in G} is subset of N for any y. Hence {yNy^-1 : y is in G} = N for any y in G. (for any z in N y(y^-1.z.y)y^-1 = z and note that y^-1.z.y is in N as y^-1.z.y = y^-1.z.(y^-1)^-1. Thus the above is not just subset but that they are equal.In fact this is a well established fact that to show N is normal you can do either yNy^-1 is equal to N for all y in G or that yNy^-1 is a subset of N for all y in G.These two are equivalent for N subgroup) So we get yNy^-1 = N for any y in G. Hence N is normal.

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