# Problem 1. Let G be a group and let H, K be two subgroups of G

advertisement ```Problem 1. Let G be a group and let H, K be two subgroups of G. Define 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, T 2 , T 4 , T 6 }? 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 −1 hk). Since k ∈ K and k −1 hk ∈ 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 &middot; e ∈ HK. If a, b ∈ HK
then a = hk and b = h1 k1 for some h, h1 ∈ H and k, k1 ∈ K. Thus ab = hkh1 k1 .
Since HK = KH and kh1 ∈ KH, we have kh1 = h2 k2 for some k2 ∈ K, h2 ∈ H.
Consequently,
ab = h(kh1 )k1 = h(h2 k2 )k1 = (hh2 )(k2 k1 ) ∈ HK
(since hh2 ∈ H and k2 k1 ∈ K). Thus HK is closed under multiplication. Finally,
1
a−1 = (hk)−1 = k −1 h−1 ∈ KH. Since KH = HK, we see that HK is closed under
taking inverses so it is a subgroup.
Note. We in fact proved that whenever HK = KH then HK is a subgroup. Two
subgroups H, K are called permutable if HK = KH. The first part of c) says
that a normal subgroup is permutable with any other subgroup.
d) Let g ∈ G and x ∈ HK. Then x = hk for some h ∈ H and k ∈ K. Thus
gxg −1 = g(hk)g −1 = (ghg −1 )(gkg −1 ) ∈ HK (since ghg −1 ∈ H and gkg −1 ∈ K).
This proves that HK is a normal subgroup.
e) By the very definition,
HK = {I, T, T 2 , T 4 , T 6 , S, ST 2 , ST 4 , ST 6 }.
Note that if we think of D16 as of symmetries of the regular 8−gon A1 A2 ...A8 (with
T (Ai ) = Ai+1 and S being the reflection in the axis passing through A1 ) then
HK consists of those symmetries which take the square A1 A3 A5 A7 into itself. In
particular HK is a subgroup of D16 . This also follows from c) since K is a normal
subgroup of D16 (prove it!). Note that HK can be identified with D8 (how ?).
Problem 2. Let G be the group from Problem 2 of Homework 31. Let H = {Ta,b ∈
G : a is a rational number}. Prove that H is a normal subgroup of G.
Solutuion: First note that H is indeed a subgroup. In fact, e = T1,0 ∈ H.
Recall that Ta,b Tc,d = Tac,ad+b If Ta,b , Tc,d ∈ H then a, c ∈ Q so ac is rational and
−1
−1
therefore Ta,b Tc,d ∈ H. Also, Ta,b
= Ta−1 ,−a−1 b so Ta,b
∈ H (since a−1 is rational).
This shows that H is a subgroup. Now if Ta,b ∈ G and Tc,d ∈ H then
−1
Ta,b Tc,d Ta,b
= Tac,ad+b Ta−1 ,−a−1 b = Taca−1 ,ac(−a−1 b)+ad+b = Tc,−cb+ad+b ∈ H
since c is rational. Thus H is a normal subgroup.
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