Show that $f$ is an homomorphism

Question

$a_1= i , a_2=(123) , a_3=(132) ,a_4=(23) , a_5=(12) , a_6=(13)$

define $f$: $S_{3} \rightarrow Z_{2}$ By $f(a_i)=0$ for $i = 1,2,3$ and $f(a_i)=1$ for $i = 4,5,6$

Do you need to prove all elements of $S_{3}$ Or 1 element example ${i}$ enough ?

Answer 1

In theory, you do need to check all the $36$ different products to see that $f$ is a homomorphism. In practice, you can do a few shortcuts letting you take care of several of the products at once. For instance, taking care of all six products $$(123)(12), (231)(23), (312)(31), (132)(13), (213)(21), (321)(32)$$ in one go by calculating generally $$f((abc)(ab)) = f((ac)) = 1\\f((abc))+f((ab)) = 0+1 = 1$$This homomorphism (and its generalisations to $S_n$) give rise to the notion of the parity of a permutation (alternatively called the sign of a permutation if $\Bbb Z_2$ is swapped for the group $\{1, -1\}$ wiht multiplication).

Answer 2

Observe that $a_{i}$ is even for $i=1,2,3$ and odd for $i=4,5,6$. Thus $f$ is in fact $\rm sgn$ (with the correspondence between $\left(\mathbb{Z}_{2},+\right)$ and $\left(\left\{\pm 1\right\},\cdot\right)$).