7: Isomorphism of Groups
( \newcommand{\kernel}{\mathrm{null}\,}\)
Two groups may look different yet be essentially the same. This concept of sameness is formalized in mathematics by the concept of isomorphism (from the Greek: isos meaning the same and morphe meaning form). Before we give a precise definition of isomorphism, let’s look at some small groups and see if we can see whether or not they meet our intuitive notion of sameness.
Problem 7.1 Go through the examples of groups we have covered so far and make a list of all those with order ≤12. List them according to their orders. For example, the following groups have order 6: GL(2,Z2),Z6,S3,U7,U9,Z2×Z3. Make a similar list for all integers from 1 to 12.
Let G={g1,g2,…,gn}. Let o(gi)=ki for i=1,2,…,n. We say that the sequence (k1,k2,…,km) is the order sequence of the group G. To make the sequence unique we assume that the elements are ordered so that k1≤k2≤⋯≤kn.
For example, the order sequence of S3 is (1,2,2,2,3,3).
Problem 7.2 Consider the following list of properties that may be used to distinguish groups.
- The order of the group.
- The order sequence of the group.
- Whether the group is abelian or not.
Look carefully at the groups in the list you made for the previous problem and see which may be distinguished by one or more of the three listed properties.
Let (G,∗) and (H,∙) be groups. A function f:G→H is said to be a homomorphism from G to H if f(a∗b)=f(a)∙f(b) for all a,b∈G. If in addition f is one-to-one and onto, f is said to be an isomorphism from G to H.
We say that G and H are isomorphic if and only if there is an isomorphism from G to H. We write G≅H to indicate that G is isomorphic to H.
Examples 7.1 Some familiar examples of homomorphisms and isomorphisms are:
- det:GL(2,R)→R−{0} is a homomorphism since det(AB)=det(A)det(B) for all A,B∈GL(2,R).
- sign:Sn→{1,−1} is a homomorphism since sign(στ)=sign(σ)sign(τ) for all σ,τ∈Sn.
- log:R+→R, where R+ denotes the positive real numbers and the operations are respectively multiplication and addition, is an isomorphism since from calculus we know that log is one-to-one and onto and log(xy)=log(x)+log(y) for all positive real numbers x and y. [Here log(x)=ln(x).]
- exp:R→R+, where R+ denotes the positive real numbers and the operations are respectively addition and multiplication, is an isomorphism since from calculus we know that exp is one-to-one and onto and exp(x+y)=exp(x)exp(y) for all real numbers x and y. Note that exp(x) is an alternative notation for ex.
The last two examples show that the group of positive real numbers under multiplication is isomorphic to the group of all real numbers under addition.
If G, H and K are groups then
- G≅G,
- If G≅H then H≅G, and
- If G≅H and H≅K, then G≅K.
Problem 7.3 Prove Theorem 7.1.
Problem 7.4 Prove that every group of order 1 is isomorphic to the group U2.
Problem 7.5 Prove that every group of order 2 is isomorphic to the group Z2.
Problem 7.6 Prove that every group of order 3 is isomorphic to the group Z3.
Problem 7.7 Prove that if G and H are isomorphic groups then |G|=|H|.
Problem 7.8 Prove that if G and H are groups then G×H≅H×G.
Let (G,∗) and (H,∙) be groups and let f:G→H be a homomorphism. Let eG denote the identity of G and, eH denote the identity of H. Then
- f(eG)=eH,
- f(a−1)=f(a)−1, and
- f(an)=f(a)n for all n∈Z.
Problem 7.9 Prove parts 1 and 2 of Theorem 7.2.
Problem 7.10 Prove part 3 of Theorem 7.2 for n=2,−2,3,−3.
The general case of Theorem 7.2 can be proved by induction on n. We will come back to this later.
Problem 7.11 Restate Theorem 7.2 (a) in the case that both G and H use additive notation, (b) in the case where G uses additive notation and H uses multiplicative notation, and (c) in the case where G uses multiplicative notation and H uses additive notation.
Let (G,∗) and (H,∙) be groups and let f:G→H be an isomorphism. Then o(a)=o(f(a)) for all a∈G. It follows that G and H have the same number of elements of each possible order.
Problem 7.12 Prove Theorem 7.3. Hint: Use the Theorem 2.
If G and H are isomorphic groups, and G is abelian, then so is H.
Problem 7.13 Prove Theorem 7.4.
A group G is cyclic if there is an element a∈G such that ⟨a⟩=G. If ⟨a⟩=G then we say that a is a generator for G.
Problem 7.14 Find an example of a cyclic group that has more than one generator.
If G and H are isomorphic groups and G is cyclic then H is cyclic.
Problem 7.15 Prove Theorem 7.5.
Problem 7.16 Determine which of the following groups are cyclic and which are not cyclic.
- Z under ordinary addition.
- Zn under addition modulo n.
- Un for n≤12.
- S3.
- Z2×Z3.
- Z2×Z2.
- Z3×Z5.
- A3.
- S4.
- GL(2,Z2).
Problem 7.17 [Challenge Problem] Prove that if G is a finite cyclic group of order n then G has ϕ(n) generators. Hint: Let G=⟨a⟩. Show than an element b=ai∈G is a generator of G if and only if gcd(i,n)=1.
Let a be an element of a group G.
- If o(a)=∞ then ⟨a⟩≅Z.
- If o(a)=n∈N then ⟨a⟩≅Zn.
Proof of 1: Assume that o(a)=∞. Define the function φ:Z→⟨a⟩ by the rule φ(n)=an for n∈Z. φ is onto by definition of ⟨a⟩. To prove that φ is one-to-one let φ(n)=φ(m) for some n,m∈Z. Then an=am. If n≠m by symmetry we can assume n<m. Then e=a0=an−n=ana−n=ama−n=am−n. But n<m so m−n∈N. This contradicts the assumption that o(a)=∞. So we must have n=m. This shows that φ is one-to-one. Since φ(n+m)=an+m=anam=φ(n)φ(m) φ is a homomorphism and it follows that φ is an isomorphism. Hence Z≅⟨a⟩. By Theorem 7.1 ⟨a⟩≅Z.
Proof of 2: Assume that o(a)=n∈N. For our proof we need to introduce the following notation from Appendix C.
Let n∈N. For each a∈Z by the Division Algorithm there are unique integers q and r satisfying a=nq+r \qquad and \qquad 0≤r<n. We denote r by amodn. That is, amodn is the remainder when a is divided by n.
Now using this we can define precisely addition modulo n by the rule: a⊕b=(a+b)modn. Note that here we write a+b for addition in Z and a⊕b for addition in Zn. So to be precise, by Zn we mean the group (Zn,⊕).
Recall that Zn={0,1,2,…,n−1}. On the other hand by Theorem 4.2 since o(a)=n we have ⟨a⟩={a0,a1,…,an−1}. So the mapping φ:Zn→⟨a⟩ defined by the rule φ(i)=ai for i=0,1,2,…,n−1, is clearly one-to-one and onto. It remains to show that φ is a homomorphism. To prove this note first that i⊕j=r means that i+j=qn+r where 0≤r≤n−1. Now we have φ(i⊕j)=φ(r)=ar=ai+j−qn=aiaja−qn=aiaj(an)−q=aiaje−q=aiaje=aiaj=φ(i)φ(j). Hence φ(i⊕j)=φ(i)φ(j). That is, φ is a homomorphism. Since it is also one-to-one and onto it is an isomorphism. Hence Zn≅⟨a⟩ and by Theorem 7.1 ⟨a⟩≅Zn. ◼
Problem 7.18 Prove that if G is a cyclic group then G is isomorphic to Z or Zn.
The above shows that a group generated by one element is easily describable. What about groups that are not generated by one element but are “generated” by two (or more elements)? The following exercise introduces a notation to make precise such matters.
Problem 7.19 [Challenge Problem] Let G be a group and let S⊂G. Define ⟨S⟩ to be the subset of G whose elements have the form sϵ11sϵ22⋯sϵnn where n∈N, si∈S and ϵi=±1 for i=1,2,…,n. Prove
- ⟨S⟩ is a subgroup of G.
- ⟨S⟩ is the smallest subgroup of G that contains S, that is, if K is a subgroup of G and S⊂K then ⟨S⟩⊂K.
- Show that for n≥3 the group Sn is not cyclic, but Sn=⟨{τ,σ}⟩ where τ=(12) and σ=(12⋯n).
Note that the above problem shows that although Sn, n≥3, is not cyclic, it is generated by two elements. However, unlike the cyclic groups one can say very little about groups generated by two elements.
You may be interested in the curious fact (first discovered by Philip Hall) that (A5)19 (i.e., the direct product of 19 copies of the alternating group of degree 5) can be generated by two elements, but (A5)20 cannot. On the other hand, the group (Z2)n, that is, the direct product of n copies of Z2, requires a minimum of n generators for each positive integer n.
We state without proof the following theorem. A proof may be found, in any of the references .
If G is a finite group of order n, then there is a subgroup H of Sn such that G≅H. ◼
This makes precise the idea that every finite group is “contained” in S_n for some positive integer n. For example, the group U_8=\{ 1,3,5,7 \} is isomorphic to the subgroup H = \{ \iota, (1 \ 2), (3 \ 4), (1 \ 2)(3 \ 4) \} of S_4. Note that a group of order k may well be isomorphic to a subgroup of S_n where n < k.
Problem 7.20 Find a group of order 120 which is ismorphic to a subgroup of S_n where n < 120.