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Math Normal Group
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1. If H and K are subsets of a group define HK={ hk | h ? H and k ? K } (a) Prove that if H is a normal subgroup of a group, then HaHb=Hab, i.e the coset. Hence, if H is a normal subgroup of a group, the operation *, defined by Ha*Hb=Hab is a well defined operation on the set of right coset of H in G.(b) Prove... 顯示更多 1. If H and K are subsets of a group define HK={ hk | h ? H and k ? K } (a) Prove that if H is a normal subgroup of a group, then HaHb=Hab, i.e the coset. Hence, if H is a normal subgroup of a group, the operation *, defined by Ha*Hb=Hab is a well defined operation on the set of right coset of H in G. (b) Prove that { Ha | a ? G} together with the operation * forms a group. This group is denoted by G/H and is called the qoutient group of G modulo H. What is the identity element? The inverse of Ha? 2. In parts (a) - (c), all the qoutient groups are cyclic and therefore are isomorphic to Zn for some n. In each case, find n, and a generator of the qoutient group. (a) Z6/<[2]> (b) Z12/<[8]> (c)Z15/<[6]>
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Math Normal Group
發問:
1. If H and K are subsets of a group define HK={ hk | h ? H and k ? K } (a) Prove that if H is a normal subgroup of a group, then HaHb=Hab, i.e the coset. Hence, if H is a normal subgroup of a group, the operation *, defined by Ha*Hb=Hab is a well defined operation on the set of right coset of H in G.(b) Prove... 顯示更多 1. If H and K are subsets of a group define HK={ hk | h ? H and k ? K } (a) Prove that if H is a normal subgroup of a group, then HaHb=Hab, i.e the coset. Hence, if H is a normal subgroup of a group, the operation *, defined by Ha*Hb=Hab is a well defined operation on the set of right coset of H in G. (b) Prove that { Ha | a ? G} together with the operation * forms a group. This group is denoted by G/H and is called the qoutient group of G modulo H. What is the identity element? The inverse of Ha? 2. In parts (a) - (c), all the qoutient groups are cyclic and therefore are isomorphic to Zn for some n. In each case, find n, and a generator of the qoutient group. (a) Z6/<[2]> (b) Z12/<[8]> (c)Z15/<[6]>
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1 (a) HaHb=HHab as H is normal Now HH ={ hk | h inH and k in H} = H Therefore HaHb=Hab (b) from (a) HaHb=Hab is a right coset of H in G. H = He is the identity HHa = Ha for all a in G Ha^-1 is the inverse of Ha Hence G/H is a group. (2) (a) n=3 generator 1+ [2] (b) As [8] = [4] in Z12 n =4 generator 1+[8] (c) As [6] = [3] in Z15. n=3 generator 1+[3]其他解答:
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