MATH 817 Notes
JD Nir
[email protected]
www.math.unl.edu/∼jnir2/817.html
September 9, 2015
H≤G
H is normal iff NG (H) = G
k
{g|gHg−1 =H }
H is normal ⇔ ∀g ∈ G, gHg −1 ⊆ H.
G = GL2 (R)
g ∈ G, x ∈ G
x0 = x, x1 = gx0 g −1 , x2 = gx1 g −1 , . . .
If xi+1 ∈
/ {x0 , . . . , xi } ∀i, then X := {x0 , x1 , . . .} gives an example where gXg −1 ⊂
6=X.
theorem Assume G is cyclic and G = hxi.
a Every subgroup of G is cyclic (+ thus of the form hxm i, some m ∈ Z.)
b Say #G = ∞ hxm i = hx` i iff ` | m. In particular, hxm i = hx` i iff |m| = |`|.
c Say #G = n < ∞.
(i) hxm i = hxgcd(m,n) i
(ii) there is a bijection
1-1
{positive divisors of n} ←→ {subgroups of G}
onto
given by a 7→ hxa i
Pf: a Let H ≤ G. If H = {e}, it’s clear. If not, xm ∈ H some m 6= 0. If xm ∈ H, then x−m ∈ H
too. ∴
T := {m ∈ N | xm ∈ H} is not empty.
Let m = min element of T . I claim H = hxm i
⊆ is clear
⊇: Say x` ∈ H. Let g = gcd(`, m) = i` + jm.
Then xg = xi`+jm = (xm )j · (x` )i ∈ H.
But 1 ≤ g ≤ m ∴ g = m ⇒ ` = m · k.
⇒ x` = (xm )k ∈ hxm i
b (⇐)` | m ⇒ m = ` · j ⇒ xm = (x` )j ∈ hx` i ⇒ hxm i ⊆ hx` i
(⇒)hxm i ⊆ hx` i ⇒ xm = (x` )j = xj` ⇒ xm−j` = e.
But |x| = ∞ ∴ m − j` = 0 ⇒ ` | m.
c n = #G.
1
JD Nir
[email protected]
MATH 817
(i) G := gcd(m, n) = im + jn
xg = (xm )i · (xn )j = x(m )i , since
n = #G ⇒ |x| = n
∴ xg ∈ hxm i + so hxg i ⊆ hxm i
g | m ⇒ m = g · k ⇒ xm = (xg )k ∈ hxg i ⇒ hxm i ⊆ hxg i. X
a
(ii) Define ϕ : {pos. divisors of n} → {subgroup of G} to be ϕ(a)
 = hx i.

e

k 
1-1: Say ϕ(a) = ϕ(b). So hxa i = hxb i. Observe hxa i = xa , x2a , . . . , xn + xia 6= xja if


1 ≤ i < j ≤ na by HW.
∴ #hxa i = na
hxa i = hxb i ⇒ na = nb ⇒ a = b.
a
onto: H ⊆ G ⇒ H = hxm i, some m ∈ Z
c(i)
⇒ H = hxg i, g = gcd(m, n). Note g | n.
⇒ H = ϕ(g).
G = any group. THe collection of all subgroups of G is a poset via ⊆. In fact, this poset is a
lattice:
∀H ≤ G + K ≤ G, ∃! smallest subgroup L with H ≤ L, K ≤ L
+ ∃! largest subgroup M with M ≤ H, M ≤ K
Prop: If {Hα : α ∈ I} is any collection of subgroups of a fixed group G, then
\
Hα is a subgroup
α∈I
of G.
Pf:
• e∈
\
α∈I
• x, y ∈
Hα since e ∈ Hα ∀α
\
α∈I
Hα ⇒ x, y ∈ Hα ∀α ⇒ xy −1 ∈ Hα ∀α ⇒ xy −1 ∈
\
Hα .
α∈I
So, given H ≤ G and K ≤ G, M := H ∩ K ≤ G and if contain every subgroup of G that is
contained in H + K.
\
Given H ≤ G + K ≤ G, let L =
Nα , where {Nα | α ∈ I} is the collection of all subgroups of G
α∈I
that contains H ∪ K.
Then L ≤ G + L ≤ N if N is any subgroup of G that contains H ∪ K.
Ex G is cyclic, #G = a.
a and b ⇒ the lattice of subgroups of G is “order isomorphic” to
def
N ∪ {0} with the p.o.: a, b ∈ N ∪ {0}, a ≤ b ⇔ b | a
2
JD Nir
[email protected]
MATH 817
G = hxi
{e} = hx0 i
...
hx i hx i . . .
2
3
⊆
⊆
2 3 5 7 11 13 17 . . .
←→
hx4 i hx6 i
⊆
⊆
4 6 9 10 . . .
⊆
≥
0
hxi = G
1
3