Suppose that % is a homomorphism from a group G onto Z6(+)Z2 and
that the kernel of % has order 5. Explain why G must have normal subgroups
of orders 5, 10, 15, 20, 30 and 60.
- myisland8132Lv 710 年前最愛解答
Since Z6(+)Z2 is an abelian group, all of its subgroup should be normal subgroups. We know that the order of Z6(+)Z2 is 12 and by Lagrange's theorem, the possible orders of the subgroups of Z6(+)Z2 are 1,2,3,4,6,12. As % is a homomorphism from a group G onto Z6(+)Z2, this means that %^(-1) is a homomorphism from a group Z6(+)Z2 onto G. Also each normal subgroup K' of Z6(+)Z2 will become a normal subgroup K of G through %^(-1). Finally, we notice that % is an 5-1 mapping from G onto Z6(+)Z2 and so the orders of the normal subgroups of G are 5, 10, 15, 20, 30 and 60 respectively.