Homomorphisms. The kernel of the homomorphism ϕ:g G, denoted Ker(ϕ), is the set of elements in G that are mapped to the identity in G.

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1 10. Homomorphisms 1 Homomorphisms Isomorphisms are important in the study of groups because, being bijections, they ensure that the domain and codomain groups are of the same order, and being operation-preserving, they coordinate the operation of the domain group with the operation of the codomain group. Indeed, it is the property of being operation-preserving that allows the group properties to transfer from domain to codomain. It is thus useful to capture this operation transfer property in functions between groups even when the underlying map is not bijective. A map ϕ:g G between groups is called a homomorphism if it is operation-preserving, i.e., if ϕ(ab) =ϕ(a)ϕ(b) for all a,b G. (Note that the operation on the left side of this last equation is the operation in G while that on the right side is the operation in G.) The kernel of the homomorphism ϕ:g G, denoted Ker(ϕ), is the set of elements in G that are mapped to the identity in G.

2 10. Homomorphisms 2 Examples: Every isomorphism of groups is a homomorphism between that same pair of groups. The kernel of an isomorphism is always the trivial subgroup. The map from R * to R * given by the absolute value function is a homomorphism because xy = x y. The kernel is the subgroup 1. The map from Z to Z n given by x a x modn is a homomorphism. (Proof: use the division algorithm to write x = q 1 n + r 1, y = q 2 n + r 2 with 0 r 1,r 2 < n, and note that r 1 = x mod n and r 2 = y mod n; then x + y = (q 1 + q 2 )n + (r 1 + r 2 ), so if r 1 + r 2 < n we have that r 1 + r 2 = (x + y)modn. Otherwise, n r 1 + r 2 < 2n and so x + y = (q 1 + q 2 +1)n +(r 1 + r 2 n) with 0 r 1 + r 2 n < n, whence r 1 + r 2 n = (x + y)mod n. In any case, (x + y)modn = (x modn) +(y modn).) Its kernel is the subgroup n of Z. Let T:R n R m be a linear transformation; then it is a homomorphism of additive groups. The kernel of T is the subgroup of R n which is the null space of the transformation. The map det:gl(2,r) R * is a homomorphism (because det(ab) = det A detb) whose kernel is SL(2,R).

3 10. Homomorphisms 3 Theorem Let ϕ:g G be a homomorphism. Then 1. ϕ(e G ) = e G ; 2. ϕ(gn ) = (ϕ(g )) n for all integers n; 3. if g is finite, then ϕ(g ) divides g ; 4. Kerϕ < G; 5. ϕ(a) =ϕ(b) a Kerϕ = b Kerϕ; 6. ϕ 1 (ϕ(g)) = g Kerϕ. Proof 1. ϕ(e G ) =ϕ(e G e G ) =ϕ(e G )ϕ(e G ) e G =ϕ(e G ). 2. We proved the case n = 0 in #1. By induction, ϕ(g n+1 ) =ϕ(g n )ϕ(g ) = (ϕ(g )) n ϕ(g) = (ϕ(g)) n+1, so that the result is true for all nonnegative n. The case n = 1 follows from #1 and the fact that ϕ(g )ϕ(g 1 ) =ϕ(e G ) =ϕ(g 1 )ϕ(g). Finally, for n < 1, ϕ(g n ) =ϕ(g n ) = (ϕ(g 1 )) n = (ϕ(g)) n = (ϕ(g )) n. 3. This follows from the observation that g n = e ϕ(g ) n =ϕ(g n ) =ϕ(e G ) = e G. 4. Suppose a,b Kerϕ. Then ϕ(ab 1 ) =ϕ(a)ϕ(b) 1 = e G e G 1 = e G. 5. ϕ(a) =ϕ(b) ϕ(a)ϕ(b) 1 = e ϕ(ab 1 ) = e ab 1 Kerϕ a Kerϕ = b Kerϕ. 6. If h ϕ 1 (ϕ(g)), then ϕ(h) =ϕ(g) h Kerϕ = g Kerϕ so h g Kerϕ; conversely, if h g Kerϕ, then h = gk with k Kerϕ, so ϕ(h) =ϕ(gk) =ϕ(g)ϕ(k) =ϕ(g) h ϕ 1 (ϕ(g)). //

4 10. Homomorphisms 4 Theorem Let ϕ:g G be a homomorphism and suppose H < G and H <G. Then 1. ϕ(h ) = {ϕ(h) h H} <G ; 2. H cyclic ϕ(h ) cyclic; 3. H Abelian ϕ(h ) Abelian; 4. H <G ϕ(h ) <ϕ(g); 5. if Kerϕ = n, then ϕ is an n-to-one function from G onto ϕ(g); 6. if H is finite, then it is a multiple of ϕ(h ) ; 7. ϕ 1 (H ) = {h G ϕ(h) H } <G ( ϕ 1 (H ) is called the pullback of H under ϕ); 8. H normal ϕ 1 (H ) normal; 9. if Kerϕ = {e} and ϕ is onto, then ϕ is an isomorphism. Proof We omit the proofs of statements #1-3 and 7-8 as they are straightforward. This leaves: 4. H <G for every h H,g G, we have ϕ(g )ϕ(h)ϕ(g) 1 =ϕ(ghg 1 ) ϕ(h ) ϕ(h ) <ϕ(g). 5. This follows from the fact that G is partitioned into cosets of Kerϕ, all of size n, and g h Kerϕ ϕ(g ) =ϕ(h). 6. Restrict the domain and codomain of ϕ:g G to ϕ:h ϕ(h). Then by #5, if the kernel of the restricted map has order n in H and it has m cosets, the restricted map is n-to-one onto ϕ(h ). Thus H = mn and ϕ(h ) = m. 9. This is an immediate consequence of #5. //

5 10. Homomorphisms 5 Corollary Let ϕ:g G be a homomorphism. Then Kerϕ <G. // Examples: The map ϕ:z 12 Z 12 given by ϕ(x) = 3x is a homomorphism (why?) with kernel equal to the subgroup 4. Thus the map is 3-to-1; for instance, ϕ 1 (6) = {2,6,10} = Also ϕ( 2 ) = 6, and we see that 2 = 6 is a multiple of 6 = 2. Suppose ψ:z 12 Z 30 is a homomorphism. Since Z 12 = 1, the map is determined completely by the value of ψ (1) = a Z 30, i.e., ψ (x) = xa. Now a Z 30 a divides 30; but also, 12a = ψ (12) = ψ (0) = 0, so a divides 12. Thus, a divides 6, their gcd. That is, a =1, 2, 3, or 6. In turn this forces a = 0; a = 15; a = 10 or 20; or a = 5 or 25. In fact, each of these corresponding maps ψ (x) = xa yields a welldefined homomorphism from Z 12 to Z 30. The map from S n to Z 2 that carries every even permutation in S n to 0 and every odd permutation to 1, is a homomorphism. its kernel, A n, must therefore be a normal subgroup of S n.

6 10. Homomorphisms 6 The First Isomorphism Theorem [Jordan, 1870] The homomorphism ϕ:g G induces a map Φ:G /Kerϕ ϕ(g) given by g Kerϕ aϕ(g ) which is an isomorphism. Proof Φ is well-defined because we know that ϕ(a) =ϕ(b) a Kerϕ = b Kerϕ for every a,b G. In fact, this also shows that Φ is one-to-one. That it is onto is clear. It is operation-preserving because // Φ((a Kerϕ)(b Kerϕ)) = Φ(ab Kerϕ) =ϕ(ab) =ϕ(a)ϕ(b) = Φ(a Kerϕ)Φ(b Kerϕ) The homomorphism γ:g G /Kerϕ that sends g to its coset g Kerϕ with respect to the kernel of ϕ is called the natural map from G to the factor group G /Kerϕ induced by the homorphism ϕ. By virtue of this last theorem, there is a clear connection among the homomorphisms ϕ, γ and Φ. This connection is represented by the diagram G ϕ ϕ(g) γ Φ G /Kerϕ

7 10. Homomorphisms 7 The diagram of homomorphisms is said to commute because following the diagram along any path produces the same map. In the diagram above, this is evidenced by the fact that ϕ = Φγ. Another way to interpret the First Isomorphism Theorem is to say that every homomorphic image of G is isomorphic to a factor of group of G. Examples: The homomorphism from Z to Z n given by x a x modn is onto, so its image is all of Z n. Since the kernel is n, we have that Z n Z / n. The complex exponential map ε:r C * given by ε(θ ) = e iθ = cosθ +isinθ takes the additive real numbers to the multiplicative complex numbers. (On p. 207, Gallian refers to it as the wrapping function.) It is a homomorphism since ε(θ 1 +θ 2 ) = e i(θ 1 +θ 2 ) = e iθ 1 e iθ 2 =ε (θ 1 )ε (θ 2 ). The image of the map is the subgroup ε(r ) of C *, called the circle group, consisting of the complex numbers on the unit circle. Since the kernel is the subgroup 2π of R, we have that the circle group is isomorphic to R / 2π.

8 10. Homomorphisms 8 Corollary Let ϕ:g G be a homomorphism between finite groups. Then ϕ(g) divides both G and G. Proof By the theorem, ϕ(g) = G/Kerϕ, which equals G / Kerϕ, so is a divisor of G. On the other hand, ϕ(g) is a subgroup of G, so ϕ(g) must divide G. // Example: The only homorphism from Z 53 to Z 54 is the zero map because the order of the image group must divide both 53 and 54, hence is necessarily the trivial subgroup of Z 54. There is also a natural map γ from a group G to the factor group G/N induced by any normal subgroup N given by g a gn. This map is a homomorphism because γ (xy) = (xy)n = (xn )(yn ) = γ (x)γ (y), and its kernel is N itself. This proves the Theorem Every normal subgroup N of a group G is the kernel of a homomorphism of G, namely the natural map to its factor group G/N. //

9 10. Homomorphisms 9 Theorem Let H be a subgroup of G. Then the factor group N(H)/C(H) of the normalizer of H by the centralizer of H is naturally isomorphic to a subgroup of Aut(H). Proof Since the normalizer of H is the set of g G for which ghg 1 = H, the map which sends g N (H ) to the inner automorphism ϕ g of H (that is, ϕ g (h) = ghg 1 ) is a homomorphism from N(H) to Aut(H). The kernel of the map is the centralizer C(H) (consisting of the elements g G for which ghg 1 = h for all h H). The result follows from the First Isomorphism Theorem. // Theorem Let H and K be subgroups of the finite group G. Then the set HK has H K elements. H K Proof The function from H K, the set of ordered pairs of elements form H and K, to HK given by (h,k) a hk is certainly an onto function. It need not be one-to-one however, as it may turn out that (h,k) ( h, k ) while hk= h k. In fact, hk= h k k k 1 = h 1 h H K. Putting l = k k 1 = h 1 h, it follows that h = hl and k = l 1 k. That is, there are H K distinct elements l for which ( h, k ) = (hl,l 1 k) a hk. Our map from H K to HK

10 10. Homomorphisms 10 is therefore H K -to-one. This means that there are precisely H K images under the map, and H K since it is onto, the result follows. // These last two results can now be used to show the Theorem Every group of order 35 is cyclic. Proof Let G be a group of order 35. By Lagrange s Theorem, its nonidentity elements can only have orders 5, 7, or 35. If any element has order 35, then G is itself cyclic, so it has only cyclic subgroups, and we are done. So we will assume then that the nonidentity elements of G have orders 5 or 7. These 34 elements cannot all have order 5, for if x = 5, then x 2,x 3,x 4 all have order 5 as well, so that elements of order 5 come in clusters of 4, and 34 is not a multiple of 4. Similarly, not all 34 elements have order 7, for these elements come in clusters of 6 and 34 is not a multiple of 6 either. So G has elements of order 5 and elements of order 7. An element of order 7 will generate a subgroup H of order 7. If K were some other subgroup of order 7, then G would contain the subset HK, which by the last theorem, has = 49 elements. As this is larger than the size of G, there

11 10. Homomorphisms 11 can only be one subgroup H of order 7. But for any g G, ghg 1 is a subgroup of G of order 7 (why?), so we must have ghg 1 = H for all g G. So N (H ) =G. Since H is cyclic, H < C(H ) <G, so 7 divides C(H) and C(H) divides 35, whence C(H) can only be 7 or 35. That is, C(H ) = H or C(H) =G. In the latter case, C(H ) =G and everything in G commutes with everything in H. So if k is an element of G of order 5, then x = hk has order 35 (why?), a situation which we have already considered. In the former case, C(H ) = H, so N(H)/C(H) has order 35/7 = 5 and therefore Aut(H) N(H)/C(H) has order 5. But H Z 7, so Aut(H) Aut( Z 7). However, Aut( Z 7) is known to have order 6, a contradiction. This shows that G must have an element of order 35, therefore is cyclic. //