Group isomorphisms and bijective maps

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SUMMARY

The discussion centers on the concept of group isomorphisms and the preservation of the group combination law through bijective maps. An isomorphism between two groups, G1 and G2, requires that a bijection f satisfies f(ab) = f(a)f(b) for all elements a, b in G1, ensuring the structure of the groups is maintained. It is established that not all bijections preserve the group law, as demonstrated with specific examples, including the integers under addition and the group Z/2Z. Furthermore, the discussion highlights that multiple bijections can exist for a single isomorphism, emphasizing the nature of automorphisms within groups.

PREREQUISITES
  • Understanding of group theory concepts, specifically isomorphisms and homomorphisms.
  • Familiarity with the properties of bijective functions.
  • Knowledge of group operations, such as addition and multiplication in various groups.
  • Basic understanding of automorphisms and their significance in group theory.
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  • Study the definition and properties of homomorphisms in group theory.
  • Explore the concept of automorphisms and their applications in various mathematical structures.
  • Investigate examples of isomorphic groups and the conditions for their isomorphism.
  • Learn about the implications of group structure preservation in algebraic systems.
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This discussion is beneficial for students of mathematics and theoretical physics, particularly those studying group theory, as well as mathematicians interested in the foundational aspects of algebraic structures.

fa2209
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Im taking a group theory course at the moment in my third year of a theoretical physics degree. In my textbook the author says defines an isomorphism by saying that if two groups are isomorphic then their elements can be put in a one-to-one correspondence that preserves the group combination law. My question is: what exactly does "preserving the combination law" mean and would any bijection do that?

My understanding of preserving the combination law is as follows. Consider 2 groups of order k. G1 ={e1, n1,...n(k-1)} & G2 = {e2, m1,...,m(k-1)}

let i be some map between them such that i(n1)=m3, i(n2) = m1, i(n3)=m6.

say the group combination law for G1 gives the following n1*n2=n3

then for i to be an isomorphism would it have to be the case that the combination law for G2 is m3*m1=m6?

and is there then only one bijection for which this is an isomorphism between G1 and G2?
 
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What is meant under the "combination law" is that the mapping "preserves" the group structure in some way, i.e. a map which preserves it is called a homomorphism (that is, if f : G1 --> G2 is a map such that for all a, b in G1 we have f(ab)=f(a)f(b)). It need not be bijective in general, but if it happens to be, it's called an isomorphism.
 
fa2209 said:
would any bijection do that?
No. As an example, consider the set {0,1,2,3} with addition modulo 4 (for example, 3+2=1), and the bijection f defined by

f(0)=1
f(1)=2
f(2)=0
f(3)=3

f(2+1)=f(3)=3
f(2)+f(1)=0+2=2.
 
Or, for example, the bijection f : Z --> Z defined on the group of integers (with addition as the group operation) with f(x) = 2x + 3.
 
preserving the group law just means

f(xy) = f(x)f(y)

For example the group Z/2Z has two elements 0 and 1 with the group law 1 + 1 = 0

The matrix that rotates the plane 180 degrees generates a group with 2 elements, itself and the rotation of 360 degrees. Map 1 to the 180 degree rotation and zero to the identity map. This is a bijection that preserves the group law.

On the other hand you could have mapped 1 to the reflection about the y-axis to get an isomorphism of Z/2Z with another group of isometries of the plane.
 
Last edited:
radou said:
Or, for example, the bijection f : Z --> Z defined on the group of integers (with addition as the group operation) with f(x) = 2x + 3.
That function isn't surjective. For example 6 isn't in its range.

However, since the condition f(xy)=f(x)f(y) implies f(e)=e', where e and e' are the identity elements of the groups, a bijection that doesn't satisfy that condition can't be an isomorphism. This means that the f:Z→Z defined by f(x)=x+1 is a bijection that isn't an isomorphism.
 
Fredrik said:
That function isn't surjective. For example 6 isn't in its range.

However, since the condition f(xy)=f(x)f(y) implies f(e)=e', where e and e' are the identity elements of the groups, a bijection that doesn't satisfy that condition can't be an isomorphism. This means that the f:Z→Z defined by f(x)=x+1 is a bijection that isn't an isomorphism.

Ah, stupid me! Yes, to correct the example, it would be a bijection if we were talking about the rationals under addition. Of course, not homomorphic since f(0) = 3.
 
fa2209 said:
and is there then only one bijection for which this is an isomorphism between G1 and G2?

no, usually not. for example, we can take G1 = G2, and have an isomorphism of G1 with itself. such an isomorphism is called an automorphism.

for example, one important automorphism of (C,+) is complex conjugation (which is also an automorphism of the multiplicative group of non-zero complex numbers).

a vector space V is an abelian group, and any linear transformation which is invertible:

T:V-->V gives rise to an automorphism of the underlying abelian group of V.

this is because for some groups, different ways of "building the group" can result in the same set.

it doesn't matter if we create a rotation group by starting with a 90 degree counterclockwise rotation, or a 270 degree counterclockwise rotation, we wind up with the same set of 4 rotations in either scenario.

so if a group is defined by specifying some generators and relations between them, there are often many ways to do this that winds up producing the same group. because of this, groups are usually only characterized "up to isomorphism", isomorphic groups being regarded as "essentially the same". they need not actually BE the same, for example, the set of complex numbers {1,i,-1,-i} under complex multiplication, and the set of integers modulo 4 under addition modulo 4, {0,1,2,3} aren't "the same thing" but they ARE isomorphic, with the isomorphism being:

k<-->exp(ikπ/2)
 

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