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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Problem #3
(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 ...

Solution Summary

Homomorphisms are investigated. The solution is detailed and well presented.

... G| = 99 and [G : H ] = 9 then H must be a normal subgroup of G ... f : Z 7 → Z 5 is a homomorphism. ... x and y are normal subgroups of G , then we have x −1 y −1 ...

... 2 If H and K are normal subgroups of a ... In other words, has a subgroup isomorphic to , which means ... try and use the isomorphism theorem, where is a homomorphism. ...

... and there is nothing to prove; all abelian subgroups must be ... So the two subsets are equal, and the subgroup ker h ... (ii) Yes, a homomorphism proof exactly as ...

... alternating groups, permutation groups, group homomorphisms, conjugacy classes, normal subgroups, general linear ... is a normal subgroup of . ... is a homomorphism. ...

... x, of G satisfying the equation x²=e form a sub group H of ... G=Z, H=2Z, then H is a normal subgroup of Z ... But if ϕ is a homomorphism, then ϕ(gh)=(gh)^(-1)=h^(-1 ...

Proofs : Grorups, Subgroups and Indexes. ... before, G acts on the left cosets of H, and this action induces a homomorphism G --> S_n ... 3) Suppose H is a subgroup of G ...

... Normal subgroups, Second Theorem of Isomorphism, Conjugates and Cyclic Groups ... Therefore, N is a normal subgroup of G ... First, I claim that ϕ is a homomorphism. ...

... This shows how to complete a proof regarding a ... 1} Prove that SLn(k) is a normal subgroup of GLn ... Consider the homomorphism det : GLn (k ) → k . Notice that ker ...

...Proof: (a) Consider any two elements (a_i) and (b_i) in ... in P and thus P is a subgroup of the ... is surjective, according to the first group homomorphism theorem, P ...

... phi is a group homomorphism. ker(phi) is the largest normal subgroup of G. For ... Abelian Groups, Non-Abelian Groups, Isomorphisms and Subgroups are investigated. ...