Question

(1) Let G be a group and H, K be subgroups of G.
(a) Show that if H is a normal subgroup, then HK = {xy|x ? H, y ? K} is a
subgroup of G.
(b) Show that if H and K are both normal subgroups, then HK is also a normal
subgroup.
(c) Give an example of subgroups H and K such that HK is not a subgroup of G.

Answer 1

(a) To prove HK is a subgroup,first we prove that HK=KH

Let a? HK. Then a = xy for some x? H and y ? K. Note that a = xy = y(y^-1 xy). Since x ? K and y^-1 xy ? H (here we use the assumption that H is normal subgroup of G), we see that a ? KH. This shows that HK ? KH. To see the opposite inclusion, consider b ? KH, so b= yx for some x ? H and y ? K. Thus b = (yxy^-1 )y? 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 = xy and b = x₁y₁ for some x,x₁? H and y,y₁ ? K. Thus ab = xyx₁y₁. Since HK = KH and yx₁ ? KH, we have yx₁ = x₂y₂ for somey₂ ? K, x₂ ? H.

Consequently,

ab = x(yx₁)y₁ = x(x₂y₂ )y₁ = (xx₂)(y₂ y₁ )? HK (since xx₂ ? H and y₂ y₁ ? K). Thus HK is closed under multiplication. Finally,

a ^-1 = (xy) ^-1 = y^-1x^-1 ? KH. Since KH = HK, we see that HK is closed under taking inverses so it is a subgroup.

(b)

Let g ? G and a ? HK. Then a= xy for some x ? H and y ? K. Thus gag^-1 = g(xy)g -1 = (gxg^-1)(gyg^-1 ) ? HK (since gxg^-1 ? H and gyg^-1 ? K). This proves that HK is a normal subgroup.

(c).