Quotient Groups Surjective Group Homomorphisms Examples
Cosets and Lagrange's theorem If G is a group and H is a subgroup we may de ne a left coset to be a subset xH = xh h H : f j 2 g Proposition 1. Any two left cosets are either equal or disjoint. Each has cardinality H . j j Corollary 2. (Lagrange's theorem) If H is a subgroup G = H [G: H] where the index [G: H] is de ned to be the number of left cosets. j j j j
Quotient Groups Let G be a group with a normal subgroup N. (Notation: N G) The de nition of normality is: g G gNg 1 = N: P 2 ) ¡ Often useful to check (explained Monday) if g G gNg 1 N 2 ) ¡ 1 this is su cient to imply normality. Reminder: we may replace g by g¡ in the last inclusion and g G g 1Ng N 2 ) ¡ g 1Ng N N gNg 1: ¡ ) ¡ Combine this with gNg 1 N give gNg 1 = N. ¡ ¡ Some subgroups are normal, some are not. For example in S3 the subgroups of order 2 are not normal but the unique subgroup (123) = A3 is normal. If X, Y are subsets of G de ne h i XY = xy x X; y Y : f j 2 2 g We have (XY )Z = X(YZ) because both sides equal xyz x X ; y Y ; z Z : f j 2 2 2 g Theorem 3. Let G/N be the set of left cosets xN of the normal subgroup N. The product of two cosets is a coset and G/N is a group (called the quotient group). More precisely: 1 1 (xN)(yN) = xyN ; (xN)¡ = x¡ N The coset N = 1 N serves as the identity in G/N. Proof. The boxed equation is the main thing to check. The identity
1 gNg¡ = N can be written gN = Ng: To prove (xN)(yN) = xyN: xNyN = xyNN = xyN:
1 1 xNx¡ N = xx¡ N = N and it is also easy to see N (xN) = (xN)N = xN so the coset N serves as an identity. Look at the boxed equation (xN)(yN) = xyN
This means we have a map : G G/N de ned by (x) = xN. This equation shows that is a homorphism. It is su¡r!jective since by de nition every element of G/N is a coset xN = (x).
Surjective group homomorphisms
Theorem 4. Let G and Q be groups and : G Q a surjective group homomor- phism. Let ¡! K = ker( ) = x G (x) = 1 : f 2 j g This is a normal subgroup. Then G = Q K : j j j j j j Proof. De ne an equivalence relation on G: x y if (x) = (y): Since surjective, the number of equivalence classes is equal to Q . Indeed, a typical equivalence class equals j j C(q) = x G (x) = q ; q Q: f 2 j g 2 Dummit and Foote use the term ber. What I will show is that the equivalence classes are cosets of K. Let x G and let q = (x). I claim that y x if and only if y xK. In other words 2 2 C(q) = xK: To prove this, rst assume y xK, say y = xk with k K. Then 2 2 (y) = (xk) = (x) (k) = (x)1 = (x) = q so y C(q). This proves that xK C(q). Conversely assume that y C(q). Then 2 1 2 (y) = q. Let k = x¡ y. We have 1 1 1 (k) = (x¡ y) = (x)¡ q = q¡ q = 1 so k K. Thus y = xk xK proving C(q) xK. Combining this with our previous inclu2sion C(q) = xK. 2 The number Q of bers equals the number of coset xK. By Lagrange's theorem, this equals [G: Kj] ajnd G = K [G: K] = K Q : j j j j j jj j
Example
We've seen that S4 has a normal subgroup of order V , the Klein 4-group: V = 1; (12)(34); (13)(24); (14)(23) : f g We'll make a group action of S4 on a set with 3 elements. Let X = (12)(34); (13)(24); (14)(23) f g And S will act on this set by conjugation. If g S and x X then gxg 1 X. 4 2 4 2 ¡ 2 1 2 3 4 g = a b c d then 1 2 3 4 a b c d g (12)(34)g 1 = (12)(34) = (ab)(cd) ¡ a b c d 1 2 3 4 shows that conjugating an element of shape (xy)(zw) gives an element of the same shape. (I haven't defined shape but appeal to your intuition.) This shows that 1 gXg¡ X and both sets have same cardinality so they are equal. The proof that V is normal uses the same calculation. Recall from Lecture 3 1 that if I have a action of G on a set X I have a homo- morphism G Sym(X) (the set of permutations of X, denoted SX in DF). If X is as above X ¡=!3 so I get a homomorphism G S . j j ¡! 3 Lemma 5. This homomorphism is surjective.
Proof. If g G let (g) Sym(X) be the corresponding permutation. Label: 2 2 A = (12)(34); B = (13)(24); C = (14)(23): g = (12)
1 gAg¡ = (12)(12)(34)(12) = (12)(34) = A
1 gBg¡ = (12)(13)(24)(12) = (14)(23) = C
1 gCg¡ = (12)(14)(23)(12) = (13)(24) = B So ((12)) = (BC): Similarly (13) = (AC). Now Sym(X) = S3 is generated by the two transpositions (BC) and (AC). Since (G) contains a set of generators of S3 it equals S3 so is surjective. By the theorem, G = ker( ) S , in other words 24 = ker( ) 6 and so j j j j j 3j j j ker( ) = 4: j j Observe that V is abelian so it acts trivially by conjugation on X V . Therefore V ker( ). Both have order 4 and so V = ker(y).