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8.2.3.a
Z(3)xZ(4),Z(3)xZ(2)xZ(2).

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8.2.3.c
Z(2)Z(3)Z(5).

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8.2.5.a
250=2*5³, therefore the elementary divisors of Z(250) are 2 and 5³.

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8.2.17.a
Suppose G and H are not isomorphic. Then there is a prime p and integers n,m,k such that in the set of the elementary divisors of G there are m copies of pⁿ while in the set of the elementary divisors of H there are k copies of pⁿ, m!=k. Then there will be 2m and 2k copies of pⁿ in the set of the elementary divisors of GxG and HxH. But we know know that GxG and HxH are isomorphic, therefore 2m=2k, which contradicts to m!=k.

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8.2.23
There are primes p(1),..., p(k) and integers a(1),...,a(k) such that G=Z(p(1)^a(1))...Z(p(k)^a(k)). If all the p(i)'s are different, than G is isomorphic to Z(p(1)^a(1))*...*p(k)^a(k)) and so it is cyclic. Suppose that G is not cyclic. Then there is a prime p and integers m,n such that G=Z(pⁿ)xZ(p^m)xZ(*)...Z(*). WLOG n=max(n,m). Now, every element in Z(pⁿ) has order which divides pⁿ, and every element in Z(p^m) has order which divides p^m which divides pⁿ, therefore every element of the form (a,0,0,...) or (0,b,0,...) in G has order that divides pⁿ. But the number of such elements is pⁿ+p^m-1 which is bigger than pⁿ. This contradicts to the hypothesis of the problem.

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Write a proof that the group of rotations of an icosahedron is isomorphic to the alternating group on 5 letters.

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