All about a cosets and Lagrange's theorem

Cosets

1. Cosets

2. Lagrange's Theorem
•The most important single theorem in
group theory. It helps answer:
–How large is the symmetry group of a
volleyball? A soccer ball?
–How many groups of order 2p where p is
prime? (4, 6, 10, 14, 22, 26, …)
–Is 2
257
-1 prime?
–Is computer security possible?
–etc.

3. Cosets
•Definition:
Let H ≤ G and a in G.
The left coset of H containing ais the set
aH = {ah | h in H}
The right coset of H containing ais the set
Ha = {ha | h in H}
•In additive groups use a+H and H+a.
•a is called the coset representativeof aH.
•Similarly, aHa
-1
={aha
-1
| h in H}

4. Cosets of K = {R
0, V} in D
4
•Some left cosets of K:
DK = {DR
0, DV} = {D,R
90}
R
180K = {R
180R
0, R
180V} = {R
180, H}
•Some right cosets of K:
KD = {R
0D, VD} = {D, R
270}
KR
180= {R
0R
180,VR
180} = {R
180, H}

5. Cosets of K = {R
0, V} in D
4
•Left Cosets
R
0K = {R
0, V} = VK
R
90K = {R
90, D} = DK
R
180K = {R
180, H} = HK
R
270K = {R
270, D'}= D'K
•Right Cosets
KR
0= {R
0, V} = KV
KR
90= {R
90, D'} = KD'
KR
180= {R
180, H} = KH
KR
270= {R
270, D} = KD

6. More cosets
•Vectors under addition are a group:
(a,b) + (c,d) = (a+c,b+d)
Identity is (0,0)
Inverse of (a,b) is (-a,-b)
Associativity is easy to verify.
•H = {(2t,t) | t in R} is a subgroup.
Proof: (2a,a) -(2b,b) = (2(a-b),a-b)

7. Visualizing H={(2t,t)}
•Let x = 2t, y = t
•Eliminate t:
y = x/2
H

8. Cosets of H={(2t,t) | t in R}
•(a,b) + H = {(a+2t,b+t)}
Set x = a+2t, y = b+t and eliminate t:
y = b + (x-a)/2
The subgroup H is the line y = x/2.
The cosets are lines parallel to y = x/2 !

9. H and some cosets
H
(1,0) + H
(–3,0)+H
(0,1) + H

10. Left Cosets of <(123)> in A
4
Let H = <(123)> {, (123), (132)}
H = {, (123), (132)}
(12)(34)H = {(12)(34), (243), (143)}
(13)(24)H = {(13)(24), (142), (234)}
(14)(23)H = {(14)(23), (134), (124)}

11. Properties of Cosets:
•Let H be a subgroup of G, and a,b in G.
•1. a belongs to aH
•2. aH = H iff a belongs to H
•3. aH = bH iff a belongs to bH
•4. aH and bH are either equal or disjoint
•5. aH = bH iff a
-1
b belongs to H
•6. |aH| = |bH|
•7. aH = Ha iff H = aHa
-1
•8. aH ≤ G iff a belongs to H

12. 1. a belongs to aH
•Proof: a = ae belongs to aH.

13. 2. aH=H iff a in H
•Proof: (=>) Given aH = H.
By (1), a is in aH = H.
(<=) Given a belongs to H. Then
(i) aH is contained in H by closure.
(ii) Choose any h in H.
Note that a
-1
is in H since a is.
Let b = a
-1
h. Note that b is in H. So
h = (aa
-1
)h = a(a
-1
h) = ab is in aH
It follows that H is contained in aH
By (i) and (ii), aH = H

14. 3. aH = bH iff a in bH
•Proof: (=>) Suppose aH = bH. Then
a = ae in aH = bH.
(<=) Suppose a is in bH. Then
a = bh for some h in H.
so aH = (bh)H = b(hH) = bH by (2).

15. 4. aH and bH are either
disjoint or equal.
•Proof: Suppose aH and bH are not
disjoint. Say x is in the intersection of
aH and bH.
Then aH = xH = bH by (3).
Consequently, aH and bH are either
disjoint or equal, as required.

16. 5. aH = bH iff a
-1
b in H
•Proof:aH = bH
<=> b in aH by (3)
<=> b = ah for some h in H
<=> a
-1
b = h for some h in H
<=> a
-1
b in H

17. 6. |aH| = |bH|
•Proof: Let ø: aH –>bH be given by
ø(ah) = bh for all h in H.
We claim ø is one to one and onto.
If ø(ah
1) = ø(ah
2), then bh
1= bh
2
so h
1= h
2. Therefore ah
1= ah
2.
Hence ø is one-to-one.
ø is clearly onto.
It follows that |aH| = |bH| as required.

18. aH = Ha iff H = aHa
-1
•Proof: aH = Ha
<=> each ah = h'a for some h' in H
<=> aha
-1
= h' for some h' in H
<=> H = aHa
-1
.

19. aH≤G iff a in H
•Proof: (=>) Suppose aH ≤ G.
Then e in aH.
But e in eH, so eH and aH are not
disjoint. By (4), aH = eH = H.
(<=) Suppose a in H.
Then aH = H ≤ G.