Let h be a subgroup of a group g. we call h characteristic in g if for any automorphism σ∈aut(g) of g, we have σ(h)=h.
(a) prove that if σ(h)⊂h for all σ∈aut(g), then h is characteristic in g.
(b) prove that the center z(g) of g is characteristic in g.

Problem 1. Let G be a group and let H, K be two subgroups of G. Dene the set HK = {hk : h ∈ H,k ∈ K}.

a) Prove that if both H and K are normal then H ∩ K is also a normal subgroup of G.

b) Prove that if H is normal then H ∩ K is a normal subgroup of K.

c) Prove that if H is normal then HK = KH and HK is a subgroup of G.

d) Prove that if both H and K are normal then HK is a normal subgroup of G.

e) What is HK when G = D16, H = {I,S}, K = {I,T2,T4,T6}? Can you give geometric description of HK?

Solution: a) We know that H ∩ K is a subgroup (Problem 3a) of homework 33). In order to prove that it is a normal subgroup let g ∈ G and h ∈ H ∩ K. Thus h ∈ H and h ∈ K. Since both H and K are normal, we have ghg−1 ∈ H and ghg−1 ∈ K. Consequently, ghg−1 ∈ H ∩ K, which proves that H ∩ K is a normal subgroup.

b) Suppose that H G. Let K ∈ k and h ∈ H ∩ K. Then khk−1 ∈ H (since H is normal in G) and khk−1 ∈ K (since both h and k are in K), so khk−1 ∈ H ∩ K. This proves that H ∩ K K.

c) Let x ∈ HK. Then x = hk for some h ∈ H and k ∈ K. Note that x = hk = k(k−1hk). Since k ∈ K and k−1hk ∈ H (here we use the assumption that H G), we see that x ∈ KH. This shows that HK ⊆ KH. To see the opposite inclusion, consider y ∈ KH, so y = kh for some h ∈ H and k ∈ K. Thus y = (khk−1)k ∈ HK, which proves that KH ⊆ HK and therefoere HK = KH. To prove that HK is a subgroup note that e = e · e ∈ HK. If a,b ∈ HK then a = hk and b = h1k1 for some h,h1 ∈ H and k,k1 ∈ K. Thus ab = hkh1k1. Since HK = KH and kh1 ∈ KH, we have kh1 = h2k2 for some k2 ∈ K, h2 ∈ H. Consequently,

ab = h(kh1)k1 = h(h2k2)k1 = (hh2)(k2k1) ∈ HK

(since hh2 ∈ H and k2k1 ∈ K). Thus HK is closed under multiplication. Finally,

Step-by-step explanation:

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.