Could you explain how to show 5.30 in detail? I also included lists of definitions and theorems in the book as a reference. 5.22. Find the invariant factors for each of the following groups.1. Z3 × Z3 × Z, × Z25 × Z11 × Z1212. Z4 × Zg × Z; × Z16 × Z5 × Z25 × Z495.30. Let G be a finite abelian group. Suppose that, for every n e N, there are atmost n elements a e G satisfying a" = e. Show that G is cyclic. Definition 5.4. Let G be a nontrivial finite abeliangroup,andsaythat GH2 x ... x Hk, where each H; is cyclic of order p", for some prime p; and positiveinteger n;. Then the elementary divisors of G are the numbers p"', p"², .. , p*where the order in this list is irrelevant, but each number must be listed as manytimes as it occurs. The trivial group has no elementary divisors.Definition 5.5. Let G be an abelian group and n a positive integer. Then we writeG" = {a" : a e G}.Lemma 5.7. Let G and H be abelian groups and n a positive integer. Then1. G" is a subgroup of G; and.2. if a : G → H is an onto homomorphism, then a(G")= H".Theorem 5.6. Let G and H be finite abelian groups. Then G and H are isomorphicif and only if they have the same elementary divisors.Theorem 5.7 (Invariant Factor Decomposition). Suppose that G is a nontrivialfinite abelian group. Then G = Hị × H2 x ..x Hk, where each H; is a cyclicsubgroup of G of order m;, with m1 1 and m;|m;+1, for 1 1 and m¡|m¡+1,mk are called the invariant factors of G.Theorem 5.8. Let G and H be nontrivial finite abelian groups. Then G and H areisomorphic if and only if they have the same invariant factors.Example 5.10. The elementary divisors of Z, × Z9 × Z3 × Z125 are 9, 9, 3, 125.Example 5.11. To find the elementary divisors of Z300 × Z3, we use Theorem 5.4 tosee that the group is isomorphic to Z25 × Z4 × Z3 × Z3, so the elementary divisorsare 4, 3, 3, 25.Example 5.12. The five abelian groups of order 16 listed in Example 5.7 are all non-isomorphic, as they have different elementary divisors. Similarly for the six abeliangroups of order 200 given in Example 5.9.Example 5.13. Let G = Z200 × Zg × Z6, H = Z120 × Z10 × Z4 × Z2 and K =Z25 x Z24 x Z; × Z2. These are all abelian groups of order 9600. However, usingTheorem 5.4, we see that G is isomorphic to Zg × Z25 × Zg × Z3 × Z2, so itselementary divisors are 8, 8, 2, 3, 25. Similarly, H is isomorphic to Z3 × Zg × Z5 xZ5 x Z2x Z4 xZ2, so its elementary divisors are 8, 4, 2, 2, 3, 5, 5 and K is isomorphicto Z25 x Z3 × Zg × Zg × Z2, so its elementary divisors are 8, 8, 2, 3, 25. Therefore,G and K are isomorphic, but H is not isomorphic to either of them.6.Example 5.14. Let us use our work in Example 5.9 to find the invariant factors of theabelian groups of order 200. We apply the method from Theorem 5.7. ConsideringZ4 x Z2 x Z25, we see that the highest power of 2 that appears is 4, and the highestpower of 5 is 25. Therefore, m = 4.25 = 100. Deleting Z4 and Z25, we are left withZ2, so mk-1 = 2, and we are finished. Thus, our group is isomorphic to Z2 × Z100,so the invariant factors are 2, 100. When we examine Z2 x Z2 x Z2 × Z5 x Zs, wesee that mk =2·5 = 10. Deleting Z2 and Z5, we are left with Z2 x Z2 x Z5. Thus,mk-1 = 2 ·5 = 10. Deleting Z2 and Z5, we are left only with Z2. Thus, mk-2 = 2,and we are finished. Therefore, our group is isomorphic to Z2 × Z10 × Z10, whichgives invariant factors of 2, 10, 10. Considering Zg × Z25, we simply get Z200, so 200is the only invariant factor. Looking at Z2 × Z2 × Z2 × Z25, we have Z2 × Z2 × Z50,so the invariant factors are 2, 2, 50. When we examine Zg x Z5 x Z5, we obtainZ5 x Z40, so the invariant factors are 5, 40. Finally, if we take Z4 x Z2 x Z5 x Z5,then we get Z10 × Z20, so the invariant factors are 10, 20.

Question

Could you explain how to show 5.30 in detail? I also included lists of definitions and theorems in the book as a reference. 5.22. Find the invariant factors for each of the following groups.1. Z3 × Z3 × Z, × Z25 × Z11 × Z1212. Z4 × Zg × Z; × Z16 × Z5 × Z25 × Z495.30. Let G be a finite abelian group. Suppose that, for every n e N, there are atmost n elements a e G satisfying a&#34; = e. Show that G is cyclic. Definition 5.4. Let G be a nontrivial finite abeliangroup,andsaythat GH2 x ... x Hk, where each H; is cyclic of order p&#34;, for some prime p; and positiveinteger n;. Then the elementary divisors of G are the numbers p&#34;', p&#34;², .. , p*where the order in this list is irrelevant, but each number must be listed as manytimes as it occurs. The trivial group has no elementary divisors.Definition 5.5. Let G be an abelian group and n a positive integer. Then we writeG&#34; = {a&#34; : a e G}.Lemma 5.7. Let G and H be abelian groups and n a positive integer. Then1. G&#34; is a subgroup of G; and.2. if a : G → H is an onto homomorphism, then a(G&#34;)= H&#34;.Theorem 5.6. Let G and H be finite abelian groups. Then G and H are isomorphicif and only if they have the same elementary divisors.Theorem 5.7 (Invariant Factor Decomposition). Suppose that G is a nontrivialfinite abelian group. Then G = Hị × H2 x ..x Hk, where each H; is a cyclicsubgroup of G of order m;, with m1 > 1 and m;|m;+1, for 1 1 and m¡|m¡+1,mk are called the invariant factors of G.Theorem 5.8. Let G and H be nontrivial finite abelian groups. Then G and H areisomorphic if and only if they have the same invariant factors.Example 5.10. The elementary divisors of Z, × Z9 × Z3 × Z125 are 9, 9, 3, 125.Example 5.11. To find the elementary divisors of Z300 × Z3, we use Theorem 5.4 tosee that the group is isomorphic to Z25 × Z4 × Z3 × Z3, so the elementary divisorsare 4, 3, 3, 25.Example 5.12. The five abelian groups of order 16 listed in Example 5.7 are all non-isomorphic, as they have different elementary divisors. Similarly for the six abeliangroups of order 200 given in Example 5.9.Example 5.13. Let G = Z200 × Zg × Z6, H = Z120 × Z10 × Z4 × Z2 and K =Z25 x Z24 x Z; × Z2. These are all abelian groups of order 9600. However, usingTheorem 5.4, we see that G is isomorphic to Zg × Z25 × Zg × Z3 × Z2, so itselementary divisors are 8, 8, 2, 3, 25. Similarly, H is isomorphic to Z3 × Zg × Z5 xZ5 x Z2x Z4 xZ2, so its elementary divisors are 8, 4, 2, 2, 3, 5, 5 and K is isomorphicto Z25 x Z3 × Zg × Zg × Z2, so its elementary divisors are 8, 8, 2, 3, 25. Therefore,G and K are isomorphic, but H is not isomorphic to either of them.6.Example 5.14. Let us use our work in Example 5.9 to find the invariant factors of theabelian groups of order 200. We apply the method from Theorem 5.7. ConsideringZ4 x Z2 x Z25, we see that the highest power of 2 that appears is 4, and the highestpower of 5 is 25. Therefore, m = 4.25 = 100. Deleting Z4 and Z25, we are left withZ2, so mk-1 = 2, and we are finished. Thus, our group is isomorphic to Z2 × Z100,so the invariant factors are 2, 100. When we examine Z2 x Z2 x Z2 × Z5 x Zs, wesee that mk =2·5 = 10. Deleting Z2 and Z5, we are left with Z2 x Z2 x Z5. Thus,mk-1 = 2 ·5 = 10. Deleting Z2 and Z5, we are left only with Z2. Thus, mk-2 = 2,and we are finished. Therefore, our group is isomorphic to Z2 × Z10 × Z10, whichgives invariant factors of 2, 10, 10. Considering Zg × Z25, we simply get Z200, so 200is the only invariant factor. Looking at Z2 × Z2 × Z2 × Z25, we have Z2 × Z2 × Z50,so the invariant factors are 2, 2, 50. When we examine Zg x Z5 x Z5, we obtainZ5 x Z40, so the invariant factors are 5, 40. Finally, if we take Z4 x Z2 x Z5 x Z5,then we get Z10 × Z20, so the invariant factors are 10, 20.

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5.30. Let G be a finite abelian group. Suppose that, for every n e N, there are at most n elements a e G satisfying a" = e. Show that G is cyclic.