Group Isomorphisms - Some Intuition

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Barbra Ryan
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1 Group Isomorphisms - Some Intuition The idea of an isomorphism is central to algebra. It s our version of equality - two objects are considered isomorphic if they are essentially the same. Before studying the technical definitions, consider a (slightly mathematical) example from language. We ll define a few finite sets: 1. { I, II, III, IV, V, VI } 2. { uno, dos, tres, cuatro, cinco, seis } 3. { 一, 二, 三, 四, 五, 六 } 4. { one, two, three, four, five, six } 5. { 1, 2, 3, 4, 5, 6 } 6. { un, deux, trois, quatre, cinq, six } 7. { I, II, III, IV, V, VI } Even if you don t recognize all of them, hopefully at least one or two sets look familiar. Based on that, you might make a conjecture about a set you don t recognize. They are, in fact, all the same - the first element of each set is used to represent the idea of a single entity, or the first object in an arrangement. The second element of each set represents a pair, and so on. The fact that each set has the same number of elements is not the important part; rather, it s that we can replace one set with another, without changing the inherent meaning. Regardless of the language or symbols we use for them, numbers are the same - the number systems we use are isomorphic, and the map between them is the isomorphism (in this case, a multi-language dictionary would make a good map). There is only one set of numbers, we just have many names for them. Many groups we ve already seen are isomorphic to each other. This is a stronger property than simply counting the number of elements in the group. If two groups G and H are isomorphic, then for every element of G there is an element of H that behaves the same way: it will have the same order, among many other properties. In addition, the groups themselves will have similar properties, even though their elements and operations are different. Even when groups appear to be similar (perhaps they have the same order, or are both infinite and Abelian, etc) they may not be isomorphic.

2 Group Isomorphisms Definition Let φ : G H be a map from a group G to a group H such that 1. φ preserves the operations of the groups (that is, φ is a homomorphism) 2. φ is onto: for every h H, there is some g G such that φ(g) = h 3. φ is one-to-one: if g 1, g 2 G and g 1 g 2, then φ(g 1 ) φ(g 2 ) If such a map exists, then G and H are called isomorphic, and the map φ is called an isomorphism. In this case, we write G H. Proving that groups are isomorphic requires finding a map between the groups, and showing that it is an operation-preserving bijection. Example We ll show that R, the group of real numbers under addition, is isomorphic to the group G of positive real numbers under multiplication. Define a map φ : R G by φ(x) = e x (or if you prefer, φ(x) = exp(x)). First note that φ is defined for any real number x, and that e x > for any x, so this really is a map from R into G. Next, we show that all three properties of an isomorphism are satisfied: 1. φ preserves the operations of the groups: Let x, y R. Then φ(x + y) = e x+y = e x e y = φ(x)φ(y) 2. φ is onto: Let g G, so g is a positive real number. Let x = ln(g). Then φ(x) = e x = e ln(g) = g 3. φ is one-to-one. Suppose x, y R such that φ(x) = φ(y). Then e x = e y, so ln(e x ) = ln(e y ) and we have x = y. Example It is important to note that a map can fail to be an isomorphism, even when it is a bijective map from a group to itself. For instance, define φ : R R by φ(x) = x 3. This is a one-to-one, onto map. However, if x, y R, we have φ(x + y) = (x + y) 3 x 3 + y 3 = φ(x) + φ(y) so the map does not preserve the group operation of addition. Theorem 1 Every infinite cycle group is isomorphic to Z under addition.

3 Cayley tables can be used to decide if two groups G and H are isomorphic. If the elements in the Cayley table of G can be relabelled with the elements of H to yield the Cayley table of H (possible after rearranging rows or columns), then G and H are isomorphic. Example Compare the table of U(5) on the left to the table of Z 4 on the right: The operations and sets are completely different, though both groups have order 4. We ll try to find a way of relabeling the elements in the table of U(5) with those of Z 4. First, the identity of U(5) is 1, while the identity of Z 4 is, so it makes sense to relabel 1 with : Next, note that the element 4 in U(5) has order 2, just like the element 2 in Z 4. So, next we ll relabel 4 with 2: Similarly, the element 3 in U(5) has order 3, just like the element 3 in Z 4, so we ll keep the 3 s where they are. While we re at it, we ll replace 2 in U(5) with the only remaining element of Z 4, 1 (both of these have order 4): This is still not exactly what we were looking for, but if we switch the last two rows and last two columns of the table (so the row and column labels are in order), we have So, the groups U(5) and Z 4 are isomorphic. If we need the specific isomorphism, it s the map that accomplishes our relabelling: φ : U(5) Z 4 defined by φ(1) =, φ(2) = 1, φ(3) = 3, φ(4) = 2 This is clearly a bijective, but as an example of how the operations are preserved: φ(2 5 3) = φ(1) = = = φ(2) + 4 φ(3)

4 The next theorem can be a very useful way of deciding if a homomorphism between groups is actually an isomorphism: we just have to find its kernel. If the kernel contains an element other than the identity, the map cannot be an isomorphism. Theorem 2 A group homomorphism is one-to-one if and only if the kernel contains only the identity element. Theorem 3 (Properties of Isomorphic Groups) Suppose G and H are isomorphic groups, with isomorphism φ : G H. Then 1. G and H have the same order. 2. If G and H are finite, they have the same number of elements of each order. 3. G is Abelian if and only if H is Abelian. 4. G is cyclic if and only if H is cyclic. If G = g, then H = φ(g). 5. φ is invertible, and φ 1 : H G is also an isomorphism. 6. If K is a subgroup of G, then φ(k) is a subgroup of H. 7. If Z(G) is the center of G, then φ(z(g)) is the center of H. Theorem 4 (Properties of Isomorphisms and Group Elements) Let φ : G H be a group isomorphism. Then 1. φ(e G ) = e H, where e G, e H are the identity elements of G, H, respectively. 2. for all g G, φ(g n ) = (φ(g)) n. 3. g = φ(g) for all g G. 4. if g 1, g 2 G, then g 1 g 2 = g 2 g 1 if and only if φ(g 1 )φ(g 2 ) = φ(g 2 )φ(g 1 ). Definition An isomorphism φ : G G from a group G to itself is called an automorphism. Theorem 5 The set of automorphisms on a group G (that is, the set of all isomorphisms from a group to itself) forms a group under function composition, denoted Aut(G). Proof Before starting the proof, it s important to understand the group we re working with. The elements of Aut(G) are isomorphisms (functions), not the elements of G. Furthermore, its operation is function composition, regardless of the operation on G. We need to show several things here. First, we need to prove that the composition of automorphisms leads to another automorphism; that is, that composition is a closed, binary operation on this set. Next, we have to show that the group properties hold. Let φ and ψ be automorphisms on the group G, so both map G into itself, both preserve group operations, and both are bijections. We have to show that φ ψ also maps from G into itself, and preserves group operations, and is a bijection. The first part is easy: if g G, then ψ(g) is also in G, which means φ(ψ(g)) is in G, so φ ψ maps elements of G to elements of G. To show φ ψ is injective, suppose there exist g 1, g 2 G such that (φ ψ)(g 1 ) = φ(ψ(g 1 )) = φ(ψ(g 2 )) = (φ ψ)(g 2 ) Since φ is an isomorphism, it is one-to-one, so ψ(g 1 ) = ψ(g 2 ). Similarly, ψ is an isomorphism, which must be one-to-one, and so g 1 = g 2. Hence, φ ψ is one-to-one. To show that φ ψ is onto, we need to show that for any g G, we can find some x G so that (φ ψ)(x) = g. Since φ and ψ are both isomorphisms, they are invertible, so we can let x = ψ 1 (φ 1 (g)). Then (φ ψ)(x) = (φ ψ)(ψ 1 (φ 1 (g))) = φ(ψ(ψ 1 (φ 1 (g)))) = φ(φ 1 (g)) = g So far, we have shown that composition is a closed, binary operation on the set of automorphisms of G. We now have to show that these automorphisms form a group under composition. Function composition is associative. To see that there is

5 an identity element, we need a map τ such that (τ φ)(x) = (φ τ)(x) = φ(x) for all automorphisms φ in the set. Let τ : G G be defined by τ(x) = x. This is an automorphism: it maps elements of G to G, and it is easy to check that the map is one-to-one, onto, and preserves the operation of the group. Moreover, τ(φ(x)) = φ(x) and φ(τ(x)) = φ(x) so this is our identity element. The last requirement is that each element of the set has an inverse. Since these are automorphisms, they are invertible, and their inverses are also automorphisms. So every automorphism in the set has an inverse (that s also in the set). Therefore, the set of automorphisms on any group G forms a group under composition. Theorem 6 Every finite cyclic group of order n is isomorphic to Z n, the integers under addition modulo n. Theorem 7 (Cayley s Theorem) Every group is isomorphic to a group of permutations. There are a few things to note about Cayley s Theorem. For one, it does not apply only to finite groups. Even infinite groups are isomorphic to some group of permutations. Also, the group of permutations does not necessary have to be a symmetric group, but simply a subgroup of some symmetric group. Every finite group of order n is isomorphic to some subgroup of S n.

Math 103 HW 9 Solutions to Selected Problems

Math 103 HW 9 Solutions to Selected Problems 4. Show that U(8) is not isomorphic to U(10). Solution: Unfortunately, the two groups have the same order: the elements are U(n) are just the coprime elements