Factor Group, Isomorphism

Question

If group $A=S_3⊕\mathbb{Z}_4$ and subgroup $B=\langle (132),2\rangle$, find a group the factor group $A/B$ is isomorphic to and construct the group table for $A/B$. I'm really not sure what to do with this one, would love to see it worked out. If I calculated properly the order of $|A|=24$ and $|B|=6$.

Answer 1

As you correctly calculated, the order of$A$ is $24$ and the order of $B$ is $6$. Therefore the quotient group is of order $4$. If you restrict the homomorphism $A\to A/B$ to the first factor, you obtain a subgroup of $A/B$ isomorphic to $\mathbb{Z}_2$, and the same is true for the second factor, which yields a different subgroup of order 2. Thus the quotient group is isomorphic to the Klein 4-group $\mathbb{Z}_2\times \mathbb{Z}_2$.

Explcitly, two generators of this group are the cosets $((12),0)B$ and $(e,1)B$. These behave like $(1,0)$ and $(0,1)$ in $\mathbb{Z}_2\times \mathbb{Z}_2$. The other two elements are $((12),1)B$, which behaves like $(1,1)$, and $(e,0)B$, which is the identity.

We have $$((12),0)B\cdot ((12),0)B=(e,1)B\cdot (e,1)B=((12),1)B\cdot ((12),1)B=(e,0)B$$ $$((12),0)B\cdot (e,1)B=(e,1)B\cdot ((12),0)B=((12),1)B$$ $$((12),1)B\cdot ((12),0)B=((12),0)B\cdot ((12),1)B=(e,1)B$$ $$((12),1)B\cdot (e,1)B=(e,1)B\cdot ((12),1)B=((12),0)B$$