Examples of Group Homomorphism

Here’s some examples of the concept of group homomorphism.

Example 1:
Let G = \left\{ {1, - 1,i, - i} \right\}, which forms a group under multiplication and I = the group of all integers under addition, prove that the mapping f from I onto G such that f\left( x \right) = {i^n}\,\,\,\forall n \in I is a homomorphism.

Solution: Since f\left( x \right) = {i^n},\,\,\,f\left( m \right) = {i^m}, for all m,n \in I

f\left( {m + n} \right) = {i^{m + n}} = {i^m} \cdot {i^n} = f\left( m \right) \cdot f\left( n \right)

Hence f is a homomorphism.

Example 2:
Show that the mapping f of the symmetric group {P_n} onto the multiplicative group G' = \left\{ {1, - 1} \right\} defined by f\left( \alpha \right) = 1 or  - 1.

According as \alpha is an even or odd permutation in {P_n} is a homomorphism of {P_n} onto G'.

Solution: We know that the product of two permutations both even or both odd is even while the product of one even and one odd permutation is odd. We shall show that

f\left( {\alpha \beta } \right) = f\left( \alpha \right)f\left( \beta \right)\,\,\,\forall \alpha, \beta \in {P_n}

(i) if \alpha, \beta are both even, then

f\left( {\alpha \beta } \right) = 1 = 1 \cdot 1 = f\left( \alpha \right) \cdot f\left( \beta \right)

(ii) if \alpha, \beta are both odd, then

f\left( {\alpha \beta } \right) = 1 = \left( { - 1} \right) \cdot \left( { - 1} \right) = f\left( \alpha \right) \cdot f\left( \beta \right)

(iii) if \alpha is odd and \beta is even, then

f\left( {\alpha \beta } \right) = - 1 = \left( { - 1} \right) \cdot 1 = f\left( \alpha \right) \cdot f\left( \beta \right)

(iv) if \alpha is even and \beta is odd, then

f\left( {\alpha \beta } \right) = - 1 = 1 \cdot \left( { - 1} \right) = f\left( \alpha \right) \cdot f\left( \beta \right)

Thus f\left( {\alpha \beta } \right) = f\left( \alpha \right)f\left( \beta \right)\,\,\,\forall \alpha, \beta \in {P_n}. Also obviously f is onto G'.

Therefore f is a homomorphism of {P_n} onto G'.

Example 3:
Show that a homomorphism from s simple group is either trivial or one-to-one.

Solution: Let G be a simple group and f be a homomorphism of G into another group G'. Then the kernel f is a normal subgroup of G. But the only normal subgroups of the simple group G are G and \left\{ e \right\}. Hence either K = G or K = \left\{ e \right\}. If K = G, the f - image of each element of G is the identity of G', as such the homomorphism f is trivial one. If K = \left\{ e \right\}, the homomorphism f is one-to-one.