# 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.