MATH 817 Notes
JD Nir
[email protected]
www.math.unl.edu/∼jnir2/817.html
August 26, 2015
I’ve posted a primilinary version of Problem Set #1, due Tuesday.
Def: G group, x ∈ G
The order of x is |x| = min {n ∈ N|xn = e} ∈ N ∪ ∞
Lemma x` = e ⇔ |x||`
Pf n = |x|. ` = q · n + r, q ∈ Z, 0 ≤ r ≤ n − 1
(⇒) e = x` = xq·n · xr = (xn )q · xr = xr
(By HW, e, x, x2 , . . . , xn−1 are distinct.)
∴ r = 0 + so n|`.
(⇐) ` = qn ⇒ x` = (xn )q = e
s
Recall D2n = symmetries of a regular n-gon = Pn
r = rotation by 2π
n radians c.c.
s = reflection about y-axis
|r| = n
srs =
r
P5 =
|s| = 2
r−1
or srsr = e
Prop #D2n = 2n + D2n = e, r, r2 , . . . , rn−1 , s, sr, . . . , srn−1
Pf: I claim e, r, . . . , rn−1 , s, sr, . . . , srn−1 are distinct.
e, r, . . . , rn−1 are distinct since |r| = n by HW.
If sri = srj , then ri = rj + so i = j. s, sr, . . . , srn−1 are distinct.
If ri = srj , then s = ri−j
So, s is a rotation.
But every reflection reverse “orientation” + every rotation preserves “orientation”. So this is not
possible.
What is sri ?
2
i=1
1
1
4
4
1
2
3
3
2
r
3
4
1
MATH 817
JD Nir
[email protected]
sr = reflection across line x = y
General rule: sri = reflection across a line of symmetry of Pn
Pn has exactly n lines of symmetry
Let g ∈ D2n . Then g is an isometry of R2 fixing (0, 0) ⇒
g = a rotation or g = a reflection.
g(Pn ) = Pn ⇒ g = ri , some i, or g = reflection across a line of symmetry of Pn .
There are n such lines + thus if g is a reflection, g ∈ s, sr, . . . , srn−1 srs = r−1 = rn−1
⇓
rs = srn−1
rs = ?
r2 s
=r·
srn−1
=
srn−1 rn−1
=
=srn ·rn−2
sr2n−2
= srn−2
ri s = srn−i
D2n = hr, s|rn = e, s2 = e, srsr = ei
G = hx1 , . . . xn |R1 = e, R2 = e, . . . , Rm = ei denotes the group generated by x1 , . . . , xn subject
only to the relations R1 = e, R2 = e, . . .
Rigorous Def: hx1 , . . . , xn |R1 , . . . , Rm i = Free group on x1 , . . . , xn
(smallest normal subgroup that contains R1 , . . . , Rm )
n=2
n=1
n=0
D4 = ?
D2 = ?
D0 = ?
X
D0 = hr, s|
r0X
=X
e, s2 = e, srsr = ei
X s2 = e
rs = sr−1
s = reflection about y-axis
r = rotation by 2πy radians, y ∈
/Q
|r| = ∞
2
JD Nir
[email protected]
MATH 817








Symmetric


Group
= Sn = σ : {1, 2, . . . , n} → {1, 2, . . . , n} |σ is 1–1 + onto
|
{z
}



on n symbols


k


bijective
·=◦
0 : The comp. of two bijections is again a bijection
1 : composition is associative
2 : e = identity function
3 : bijections are invertible.
∴ (Sn , ◦) is a group.
1 2
σ ∈ Sn , σ =
i1 i2
3
i3
...
...
n
in
σ(j) = ij {i1 , i2 , . . . , in } = {1, 2, . . . , n}
For m ≥ 2, an m-cycle in Sn is the permutation σ = (a1 a2 · · · am ) for some a1 , . . . , am ∈ 1, . . . , n
distinct given by σ(ai ) = ai+1 for 1 ≤ i ≤ m − 1
σ(am ) = a1
σ(i) = i if i ∈
/ {a1 , . . . , am }
1 2
e.g. (1 2 3) ∈ S7 is
2 3
1 2 3
(6 2 4) ∈ S8 is
1 4 3
3 4 5 6 7
1 4 5 6 7
4 5 6 7 8
6 5 2 7 8
(1 2) ◦ (2 3) = (3 1 2) 6= (2 3) ◦ (1 2) = (1 3 2)
in Sn ∀n ≥ 3
Sn , n ≥ 3, is not abelian
3