# Group isomorphisms and bijective maps

1. Dec 15, 2011

### fa2209

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?

2. Dec 15, 2011

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.

3. Dec 15, 2011

### Fredrik

Staff Emeritus
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.

4. Dec 15, 2011

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.

5. Dec 15, 2011

### lavinia

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: Dec 15, 2011
6. Dec 15, 2011

### Fredrik

Staff Emeritus
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.

7. Dec 15, 2011

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.

8. Dec 15, 2011

### Deveno

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)