# 23.1. Proof of the fundamental theorem of homomorphisms (FTH). We start by recalling the statement of FTH introduced last time.

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1 23. Quotient groups II Proof of the fundamental theorem of homomorphisms (FTH). We start by recalling the statement of FTH introduced last time. Theorem (FTH). Let G, H be groups and ϕ : G H a homomorphism. Then G/Ker ϕ = ϕ(g). ( ) Proof. Let K = Ker ϕ and define the map Φ : G/K ϕ(g) by Φ(gK) = ϕ(g) for g G. We claim that Φ is a well defined mapping and that Φ is an isomorphism. Thus we need to check the following four conditions: (i) Φ is well defined (ii) Φ is injective (iii) Φ is surjective (iv) Φ is a homomorphism For (i) we need to prove the implication g 1 K = g 2 K Φ(g 1 K) = Φ(g 2 K). So, assume that g 1 K = g 2 K for some g 1, g 2 G. Then g1 1 g 2 K by Theorem 19.2, so ϕ(g1 1 g 2) = e H (recall that K = Ker ϕ). Since ϕ(g1 1 g 2) = ϕ(g 1 ) 1 ϕ(g 2 ), we get ϕ(g 1 ) 1 ϕ(g 2 ) = e H. Thus, ϕ(g 1 ) = ϕ(g 2 ), and so Φ(g 1 K) = Φ(g 2 K), as desired. For (ii) we need to prove that Φ(g 1 K) = Φ(g 2 K) g 1 K = g 2 K. This is done by taking the argument in the proof of (i) and reversing all the implication arrows. (iii) First note that by construction Codomain(Φ) = ϕ(g). Thus, for surjectivity of Φ we need to show that Range(Φ) = Φ(G/K) is equal to ϕ(g). This is clear since Φ(G/K) = {Φ(gK) : g G} = {ϕ(g) : g G} = ϕ(g). (iv) Finally, for any g 1, g 2 G we have Φ(g 1 K g 2 K) = Φ(g 1 g 2 K) = ϕ(g 1 g 2 ) = ϕ(g 1 )ϕ(g 2 ) = Φ(g 1 K)Φ(g 2 K) where the first equality holds by the definition of product in quotient groups. Thus, Φ is a homomorphism. So, we constructed an isomorphism Φ : G/Ker ϕ ϕ(g), and thus G/Ker ϕ is isomorphic to ϕ(g). 1

2 Applications of FTH. In most applications one uses a special case of FTH stated last time as Corollary 22.5: If ϕ : G H is a surjective homomorphism, then G/Ker ϕ = H. (***) Typically this result is being applied as follows. We are given a group G, a normal subgroup K and another group H (unrelated to G), and we are asked to prove that G/K = H. By (***) to prove that G/K = H it suffices to find a surjective homomorphism ϕ : G H such that Ker ϕ = K. Example 1: Let n 2 be an integer. Prove that Z/nZ = Z n. We already established this isomorphism in Lecture 22 (see Corollary 22.3), so the point of this example is mostly to illustrate how FTH works. In this example G = Z, H = Z n and K = nz. Define the map ϕ : Z Z n by ϕ(x) = [x] n. It is straightforward to check that ϕ is a surjective homomorphism (anyway, this was verified in Lecture 15). We have Ker ϕ = {x Z : [x] n = [0] n } = {x Z : x = nk for some k Z} = nz = K. Thus, by FTH (or, more precisely, by (***)) we have Z/nZ = Z n. Example 2: Let U be the group of rotations of the unit circle in R 2. Prove that U = R/Z. Remark: As usual, by R we denote the group of reals (with addition) and Z is thought of as a subgroup of R. In this example G = R, H = U and K = Z. By definition, U = {r α : α R}, where r α is the counterclockwise rotation by α radians. Clearly, the group operation on U is given by r α r β = r α+β for all α, β R. Define the map ϕ : R U by ϕ(x) = r 2πx for all x R. Then ϕ is a homomorphism since ϕ(x)ϕ(y) = r 2πx r 2πy = r 2π(x+y) = ϕ(x + y), and ϕ is surjective, since any element of U is equal to r α for some α R, and any α R can be written as 2πx for some x R (namely x = α/2π). Finally, Ker ϕ consists of all x R such that r 2πx is the trivial rotation. But a rotation by the angle of α radians is trivial if and only if α is an integer multiple of 2π. Thus, x Ker ϕ 2πx = 2πk for some k Z x Z.

3 3 Thus, Ker ϕ = Z = K, as desired, and again by FTH we conclude that R/Z = U. Note that in this example we managed to determine the isomorphism class of the quotient group R/Z without having to visualize it. We will return to the latter problem later in this lecture. Example 3: Prove that the alternating group A n (the subgroup of even permutations in S n ) has index 2 in S n. This can be proved in a number of different ways; using FTH is just one of them. To prove that [S n : A n ] = 2 we will construct a surjective homomorphism ϕ : S n Z 2 with Ker ϕ = A n. If this is achieved, it would follow that S n /A n = Z2, so S n /A n = Z 2 = 2, and therefore [S n : A n ] = S n /A n = 2, as desired. Define ϕ : S n Z 2 by ϕ(f) = { [0] if f is even [1] if f is odd. By construction ϕ is surjective. To prove that ϕ is a homomorphism we need to show that ϕ(f) + ϕ(g) = ϕ(fg) for all f, g S n ( ) Recall (Proposition A.3 in the notes on even/odd permutations) that if f and g are both even or both odd, then fg is even if f is even and g is odd, or if f is odd and g is even, then fg is odd. Let us consider 4 cases. 1. f and g are both even. Then fg is also even. So, ϕ(f) = ϕ(g) = ϕ(fg) = [0]. Since [0] + [0] = [0], (***) holds. 2. f is even, and g is odd. Then fg is odd. So, ϕ(f)+ϕ(g) = [0]+[1] = [1] = ϕ(fg). 3. f is odd, and g is even. This case is analogous to Case f and g are both odd. Then fg is even, so ϕ(f) + ϕ(g) = [1] + [1] = [0] = ϕ(fg). Thus, we verified that ϕ is a homomorphism. Finally, Ker ϕ = {f S n : ϕ(f) = [0] 2 } is the set of all even permutations, so Ker ϕ = A n (by definition of A n ) Transversals. Definition. Let G be a group and H a subgroup of G. A subset T of G is called a transversal of H in G if T contains PRECISELY one element from each left coset with respect to H.

4 4 Example: Let G = Z and H = 3Z. Then there are 3 left cosets with respect to H: 0 + H, 1 + H and 2 + H, so the set T = {0, 1, 2} is a transversal. Another transversal is {2, 7, 9}. In general, in this example, a set T will be a transversal T = 3 and T contains one integer divisible by 3, one integer congruent to 1 mod 3 and one integer congruent to 2 mod 3. If T is a transversal of H in G, then by definition T = G/H, that is, T has the same size as the quotient set G/H. In fact, there is a natural bijective mapping T G/H given by t th. Assume now that H is normal, so that G/H is a group. Then we can define a binary operation on T so that (T, ) is a group which is isomorphic to G/H. This can be done as follows: for each g G denote by ḡ the unique element of T which lies in the coset gh. Note that ḡ = g g T. Now define a binary operation on T by setting t 1 t 2 = t 1 t 2 for all t 1, t 2 T (!!!) The following proposition is left as an exercise: Proposition (T, ) is a group, which is isomorphic to G/H via the map ι : T G/H given by ι(t) = th. We can now use Proposition 23.1 to give a new interpretation of the cylcic groups Z n and also better visualize the quotient group R/Z. Example A: Let n 2 be an integer. We already proved that the quotient group Z/nZ is isomorphic to Z n. Let G = Z, H = nz and T = {0, 1,..., n 1}. Then T is clearly a transversal of H in G, and in the above notations for any x Z we have x = the remainder of dividing x by n. Thus, by Proposition 23.1, G/H = Z/nZ is isomorphic to the following group which we denote by Z n: As a set Z n = {0, 1,..., n 1}, the set of integers from 0 to n 1. The group operation + on Z n is defined by x + y = the remainder of dividing x + y by n. From this description you can see that Z n is essentially the same group as Z n except for minor notational differences. In fact, if you were introduced to congruence classes before this course, Z n may have been defined precisely as the group Z n above. Example B: Now let G = R (with addition) and H = Z. Let T = [0, 1) = {x R : 0 x < 1} R.

5 5 We claim that T is a transversal of H in G. Indeed, the cosets with respect to H have the form x + Z, with x R, and it is easy to see that x + Z will contain precisely one element of T, namely the fractional part of x, denoted by {x}. For instance, let x = 2.1. Then x + Z = {..., 0.9, 0.1, 1.1, 2.1, 3.1,...}, and the unique number in (x + Z) T is 0.1 = {2.1}. Thus, T is a transversal of Z in R, and in the above notations for every x R we have x = {x}. Applying Proposition 23.1, we get the following conclusion: introduce the group operation + on T = [0, 1) by x + y = {x + y}. Then (T, + ) is isomorphic to R/Z. Note that the operation + on T can be more explicitly described as follows: for every x, y T we have { x + x + y if x + y < 1 y = x + y 1 if x + y 1. (we have only two case above because if x, y T, then 0 x, y < 1, so 0 x + y < 2). Let us go back to the general case. Let G be a group, H a normal subgroup, and suppose that we found a transversal T which itself is a subgroup of G. Then for any t 1, t 2 T we have t 1 t 2 T, so t 1 t 2 = t 1 t 2. Therefore, the formula (!!!) for the operation on T simplifies to t 1 t 2 = t 1 t 2. In other words, in this case the newly defined operation on T coincides with the group operation on G restricted to T. Therefore, we obtain the following useful result as a consequence of Proposition Corollary Let G be a group and H a normal subgroup of G. Assume that there exists a transversal T of H in G such that T is also a subgroup. Then the quotient group G/H is isomorphic to T (considered as a subgroup of G). We finish with two examples in the first one there will exist a transversal which is a subgroup, and in the second one there will be no such transversal. Example 1: Let G = Z 6 and H = [3] = {[0], [3]}. There are three cosets with respect to H: H = {[0], [3]}, [1]+H = {[1], [4]} and [2]+H = {[2], [5]}. The simplest possible transversal {[0], [1], [2]} is not a subgroup, but there is another one that works: T = {[0], [2], [4]} is also a transversal, and it is clearly a subgroup (e.g. because it coincides with [2], the cyclic subgroup generated by [2]). Example 2: Now let G = Z and H = 3Z. We claim that no transversal can be a subgroup here. Indeed, in this example, as we saw earlier, every

6 6 transversal has 3 elements. On the other hand, we know (see Homework#6, Problem 9) that any subgroup of Z is equal to nz for some n, and { if n 0 nz = 1 if n = 0 In particular, Z has no subgroups of order 3, so none of them could be a transversal of H = 3Z.

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