Question

(4)(20 points) (a) Show that if H and K are subgroups of an abelian group G, then HK = {hk|he H, KE K} is a subgroup of G. (b) Show that if H and K are normal subgroups of a group G, then HK is a normal subgroup of G

Answer 1

4. Q. Let H and k be Subgroups of an abelian group G with identity e and HK = {nkl hEH SHEK} Since it and k are sub groups of G we have that HCG ,Ke Gee H and eek. Thus we have that hKE G for all ht H and KEK which gives us that HKS G. In Particular, we have that eere € HK · Given W, WE HK there exists hi h'E H and K, K'E K such that w = hk and w'= hk which gives us that wW'=(hk) (h'k') = h (kn) K'. Since G is abelian we have that kh'= hk which gives us that ww'=h(hk) k' =(hn) (KK') Since H and k are subgroups of G; hin' EH and K,K'Ek we have that hh'th and KK'E K which gives us that ww's (hh') (kk') EHK

Hence HK is closed under the operation O G. Given WEHK there exists he H and kek such that we hk which gives us that wla (hk)-' - k-6-1 Since H and K are and K are subgroup and KEK we have that heH and Kile K. Since G is abelian we have that whichik which gives us that w-lah'K'E HK. Hence HK is a subgroup of G . (Proved) subgroup of G, HEH b. Let Hik be two normal Subgroup of G: Let p = HAK Then pis a sub group of G, Since the intersection of two subgroups of G is a subgroup of G. Let Þe p and XE G . Then þE H and þEK. since it is a normal subgroup of G, PGH and 2€ G2b>-TEH. Since K is a normal Subgroup of G, ÞEK and X 6G = 2px'Ek Therefore, Þe p and xe G = xpx-TEHAKEP and this proves that HAK is a normal subgroup of G. (Proved)