Here we look at some examples to see if they are morphisms, that is, does:

φ(v1) o φ(v2) = φ(v1 o v2)

where:

- φ = isomorphism mapping
- v1 and v2 = elements of group
- o = group operation

## Example 1 vector addition with mapping = shift of origin

With mapping φ between vector additions of fixed offset V_{s} so each vector is transformed by:

φ(V) = V + V_{s}

This is not an isomorphism since:

(V_{1} + V_{s}) + (V_{2} + V_{s}) ≠ (V_{1} + V_{2}) + V_{s}

however:

(V_{1} + V_{s}) + (V_{2}) = (V_{1} + V_{2}) + V_{s}

so if we treat V_{2} as an offset and don't transform it then the equation matches.

## Example 2 vector addition with mapping = scaling

With mapping φ between vector spaces of scalar multiplication by M _{} so each vector is transformed by:

φ(V) = M*V

This is an isomorphism since:

V_{1}*M + V_{2}*M = (V_{1} + V_{2}) *M

## Example 3 multiplication with mapping = scaling

With mapping φ between vector additions of fixed scalar multiplication by M _{} so each vector is transformed by:

φ(V) = M*V

This is not an isomorphism since:

V_{1}*M * V_{2}*M ≠ (V_{1} * V_{2}) *M

So what mapping φ can we use with multipication to form an isomrphism? A common form is the 'sandwich product':

φ(v) = q v q^{-1}

I have used q here to indicate a quaternion because the various types of vector product don't really have the correct properties. Use of the sandwich product in this way is discussed on this page.

so

φ(q1*q2) = q q1 q2 q^{-1} = q q1 q ^{-1} q q2 q^{-1}= φ(q1)*φ(q2) |
(since q ^{-1} q =1) |

So we can see that this sandwich product (conjugacy) has the right properties for an isomorphism