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1. Prove or disprove (a). The order of the element (2, 3) € Z6X Z15 is 5. (b). The group Z7 × Z17× Z27 × Z37 is not cyclic. X (c). There is only one cyclic group of order 2022. (d). There is an abelian isomorphic to a non-abelian group. (e). Z3 x Z9Z27! (f). If G is an abe group of order 15 and m divides 15, then G has a subgroup of order m. (g). If G is group of order 957, then G is cyclic. (h). There is non-abelian group of order 255. (i). There is a simple group of order 2021. (j). If G is an abelian group of order 72, then G has a subgroup of order 8. (k). Z4 x Z15 Z6 x Z10. (1). If g Al Alabgood = (2, (3 4 5)) € Z10 x 65, then o(g) = 15.

1. Prove or disprove (a). The order of the element (2, 3) € Z6X Z15 is 5. (b). The group Z7 × Z17× Z27 × Z37 is not cyclic. X (c). There is only one cyclic group of order 2022. (d). There is an abelian isomorphic to a non-abelian group. (e). Z3 x Z9Z27! (f). If G is an abe group of order 15 and m divides 15, then G has a subgroup of order m. (g). If G is group of order 957, then G is cyclic. (h). There is non-abelian group of order 255. (i). There is a simple group of order 2021. (j). If G is an abelian group of order 72, then G has a subgroup of order 8. (k). Z4 x Z15 Z6 x Z10. (1). If g Al Alabgood = (2, (3 4 5)) € Z10 x 65, then o(g) = 15.

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter4: More On Groups
Section4.6: Quotient Groups
Problem 27E: 27. a. Show that a cyclic group of order has a cyclic group of order as a homomorphic image. ...
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1. Prove or disprove (a). The order of the element (2, 3) € Z6X Z15 is 5. (b). The group Z7 × Z17× Z27 × Z37 is not cyclic. X (c). There is only one cyclic group of order 2022. (d). There is an abelian isomorphic to a non-abelian group. (e). Z3 x Z9Z27! (f). If G is an abe group of order 15 and m divides 15, then G has a subgroup of order m. (g). If G is group of order 957, then G is cyclic. (h). There is non-abelian group of order 255. (i). There is a simple group of order 2021. (j). If G is an abelian group of order 72, order 8. (k). Z4 x Z15 Z6 × Z10. (1). If g = (2, (3 4 5)) € Z10 x 65, then o(g) = 15. Al en G has a subgroup of Alabgood
Transcribed Image Text:1. Prove or disprove (a). The order of the element (2, 3) € Z6X Z15 is 5. (b). The group Z7 × Z17× Z27 × Z37 is not cyclic. X (c). There is only one cyclic group of order 2022. (d). There is an abelian isomorphic to a non-abelian group. (e). Z3 x Z9Z27! (f). If G is an abe group of order 15 and m divides 15, then G has a subgroup of order m. (g). If G is group of order 957, then G is cyclic. (h). There is non-abelian group of order 255. (i). There is a simple group of order 2021. (j). If G is an abelian group of order 72, order 8. (k). Z4 x Z15 Z6 × Z10. (1). If g = (2, (3 4 5)) € Z10 x 65, then o(g) = 15. Al en G has a subgroup of Alabgood
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