7 Homomorphisms and the First Isomorphism Theorem In each of our examples of factor groups, we not only computed the factor group but identified it as isomorphic to an already well-known group. Each of these examples is a special case of a very important theorem: the first isomorphism theorem. This theorem provides a universal way of defining and identifying factor groups. Moreover, it has versions applied to all manner of algebraic structures, perhaps the most famous being the rank–nullity theorem of linear algebra. In order to discuss this theorem, we need to consider two subgroups related to any group homomorphism. 7.1 Homomorphisms, Kernels and Images Definition 7.1. Let f : G ! L be a homomorphism of multiplicative groups. The kernel and image of f are the sets ker f = fg 2 G : f(g) = eLg Im f = ff(g) : g 2 Gg Note that ker f ⊆ G while Im f ⊆ L. Similar notions The image of a function is simply its range Im f = range f, so this is nothing new. You saw the concept of kernel in linear algebra. For example if A 2 M3×2(R) is a matrix, then we can define the linear map 2 3 LA : R ! R : x 7! Ax In this case, the kernel of LA is precisely the nullspace of A. Similarly, the image of LA is the column- space of A. All we are doing in this section is generalizing an old discussion from linear algebra. Theorem 7.2. Suppose that f : G ! L is a homomorphism. Then 1. f(eG) = eL 2. 8g 2 G, (f(g))−1 = f(g−1) 3. ker f / G 4. Im f ≤ L Proof. 1. Suppose that g 2 G. Then f(g) = f(geG) = f(g)f(eG) =) eL = f(eG) by the left cancellation law. 2. Suppose that g 2 G. Then −1 −1 −1 −1 eL = f(eG) = f(gg ) = f(g)f(g ) =) (f(g)) = f(g ) 3. First suppose that k1, k2 2 ker f. Then f(k1k2) = f(k1)f(k2) = eL =) k1k2 2 ker f 1 and −1 −1 −1 f(k1 ) = (f(k1)) = eL =) k1 2 ker f It follows that ker f is a subgroup of G. To see that ker f is normal, recall, let g 2 G and k 2 ker f, then −1 −1 −1 f(gkg ) = f(g)f(k)f(g) = eL =) gkg 2 ker f This is one of the conditions equaivalent to ker f / G. 4. Let f(g1), f(g2) 2 Im f. Then f(g1)f(g2) = f(g1g2) 2 Im f and −1 −1 (f(g1)) = f(g1 ) 2 Im f Thus Im f is a subgroup of L. Examples 1. Recall that the trace function tr : Mn(R) ! R is a homomorphism of additive groups. These are Abelian groups and so the kernel of tr is automatically normal without needing the above 1 Theorem. The additive group of trace-free matrices is a normal subgroup of (Mn(R), +): ker tr = fA 2 Mn(R) : tr A = 0g / Mn(R) 2. Let f : Z36 ! Z20 be defined by f(n) = 5n (mod 20). The kernel of f is the subgroup ker f = n : 5n ≡ 0 (mod 20)g = f0, 4, 8, 12, 16 / Z36 This is simply the cyclic group C5. ( 1 if s even, 3. The map sgn : Sn ! f1, −1g given by sgn(s) = is a homomorphism of −1 if s odd, groups. Here f1, −1g is a group under multiplication. Since the identity in the target group is 1, we have ker sgn = An, the alternating group of even permutations in Sn. Indeed An / Sn. × 4. det : GLn(R) ! R is a homomorphism and so ker det = SLn(R) is a normal subgroup SLn(R) / GLn(R). Compare this with example 1. All these examples should suggest an idea to you. Not only is every kernel a normal subgroup, the converse is also true: any normal subgroup is the kernel of some homomorphism. This (loosely) is the 1st isomorphism Theorem to which we will come shortly. It may seem counter-intuiutive, but the homomorphism approach actually makes calculations with factor groups easier! As the next lemma shows, there is a very easy correspondence between the cosets of the kernel of a homomorphism, and the elements of the image. 1 This kernel is often written sln(R) (for the special-linear algebra), and is very much related to the special linear group A tr A SLn(R). Indeed the expression det e = e shows that matrix exponentiation is a map exp : sln(R) ! SLn(R). 2 Lemma 7.3. Suppose that f : G ! L is a homomorphism. Then g1 ker f = g2 ker f () f(g1) = f(g2) Proof. For all g1, g2 2 G, we have −1 g1 ker f = g2 ker f () g2 g1 2 ker f −1 () f(g2 g1) = eL −1 () f(g2) f(g1) = eL () f(g1) = f(g2) . The lemma tells us there is a bijective correspondence between the factor group G and the ker f image Im f. In the next theorem, we put this to use to help us determine what can possibly be a homomorphism. Theorem 7.4. Let f : G ! L be a homomorphism. 1. If G is a finite group then Im f is a finite subgroup of L and its order divides that of G. More succinctly: jGj < ¥ =) jIm fj jGj. 2. Similarly, jLj < ¥ =) jIm fj jLj. Note that we are only assuming one of the groups to be finite in each case. Proof. 1. If G is a finite group, then Im f is a finite subgroup of L. Lemma 7.3 tells us that f(g1) = f(g2) () g1 ker f = g2 ker f whence there are precisely as many elements of Im f as there are distinct cosets of ker f in G. But this is the definition of the index: jIm fj = (G : ker f) Given that G is finite, the discussion following Lagrange’s Theorem quickly says that jGj = jIm fj · jker fj =) jIm fj jGj 2. This is immediate from Lagrange’s Theorem: Im f is a subgroup of a finite group L and so the order of Im f divides the order of L. Examples 1. Suppose that jGj = 17 and jLj = 13. How many distinct homomorphisms are there f : G ! L? If f were such a homomorphism, the Theorem says that jIm fj divides both jGj and jLj. But the only such positive integer is 1. Since the image of any homomorphism always contains the identity (f(eG) = eL), it follows that there is only one homomorphism! f must be the function defined by f(g) = eL, 8g 2 G. More generally, if gcd(jGj , jLj) = 1, then the only homomorphism f : G ! L is the trivial function f : g 7! eL. 3 2. How many homomorphisms are there f : Z4 ! S3? Again the Theorem tells us that jIm fj divides 4 = jZ4j and 6 = jS3j: thus jIm fj is either 1 or 2. If jIm fj = 2 then Im f is a subgroup of order 2 of S3. There are exactly 3 of these: fe, mig for each i = 1, 2, 3. Since a homomorphism must map the identity to the identity we therefore have the homomorphisms 001 0 e 1 001 0 e 1 001 0 e 1 B1C Bm1C B1C Bm2C B1C Bm3C f : B C 7! B C , f : B C 7! B C , f : B C 7! B C . 1 @2A @ e A 2 @2A @ e A 3 @2A @ e A 3 m1 3 m2 3 m3 Finally if Im f has one element then f is the trivial homomorphism f(z) = e, 8z 2 Z4. There are therefore 4 distinct homomorphisms. Theorem 7.5. There are exactly d = gcd(m, n) distinct homomorphisms f : Zm ! Zn, defined by n f(x) = k x (mod n) where k = 0, . , d − 1 d There are several ways of proving this: one is in the homework. We use the above discussion on the cardinality of the image. Proof. Suppose f : Zm ! Zn is a homomorphism and d = gcd(m, n). Then jIm fj d. However Im f 2 is a subgroup of Zn which is necessarily cyclic. Recalling the structure of cyclic groups we see that n must be a subgroup of the unique cyclic subgroup of order d in Zn. This is generated by d . The only possible choices for our homomorphisms are therefore3 n f (x) = k x (mod n) k d for each k = 0, 1, 2, . , d − 1. It remains only to check that these are well-defined functions. For this, note that for any j 2 Z we have, n n m n f (x + jm) = k (x + jm) ≡ k x + knj ≡ k x (mod n) (since m 2 Z) k d d d d d = fk(x) Example Suppose that f : Z12 ! Z20 is a homomorphism. Since gcd(12, 20) = 4, the image of f must be a subgroup of C4 = h5i ≤ Z20. There are four choices: f0(x) = 0, f1(x) = 5x, f2(x) = 10x, f3(x) = 15x (mod 20) Reversing the argument, we see that there are also four distinct homomorphisms y : Z20 ! Z12, namely y0(x) = 0, y1(x) = 3x, y2(x) = 6x, y3(x) = 9x (mod 12) 2 Recall that a cyclic group Zn has exactly one subgroup, which is itself cyclic, of each order d which divides n. Therefore if Im f is a subgroup of Zn, then it is cyclic.