Let G be a group and let H be a normal subgroup of G. Let m be the index of H in G (that is, the number of cosets of H). Prove that for any a we have am H.
(b) Give an example of group G, a subgroup H of index in, and an element a G such that am is not in H. (Of course, your subgroup H had better not be normal.)
(4) (a) Suppose G is a group of size 77 and H is a group of size 52. Prove that the only homomorphism from G to H is the trivial homomorphism.
(b) Give an example of a group G of size 77 and a group K of size 56 and a nontrivial homomorphism from G to IC.
Prove that if G is an abelian group and N is any normal subgroup of G then GIN is abelian.
(b) Prove that if G is any group and N is a subgroup of Z(G) such that GiN is cyclic then G must he abelian. [Note that since N is contained in the center of G, N is automatically normal in G and N is itself abelian.]
(c) Give an example of a group G and a normal subgroup N of G such that both N and G/N are abelian, hut G is not abelian.
(d) Give an example of a group G and a subgroup N of Z(G) such that GIN is abelian but G is not abelian.
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(a) Since is a normal subgroup, then is a quotient group and we have natural homomorphism , for any . Since , then . Thus . This implies that . Therefore, for any , we must have .
(b) If is not ...
Homomorphisms are investigated. The solution is detailed and well presented.