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.