Suppose$f:S_{3}\longrightarrow R^{\ast}$is Homomorphism.Then Kernal of h has

Question

Question Suppose$f:S_{3}\longrightarrow R^{\ast}$is Homomorphism.Then Kernal of h has

$\left(1\right)Always$ at most 2 elements

$\left(2\right)$$Always$ at most 3 elements

$\left(3\right)Always$ at least 3 elements

$\left(4\right)Always$ exactly 6 elements

Book Mention Option (4) is correct

My ApproachMy problem is first fundamental theorem of isomorphism.

Option $\left(4\right)Always$ exactly 6 elements$\Longrightarrow$Ker$\left(f\right)=S_{3}$

S$_{3}/ker\left(f\right)$= {Identity Element} It's order is 1

If Ker$\left(f\right)$=A$_{3}$

|S$_{3}$/ker$\left(f\right)$| =2

And order of R$^{*}$is infinite

R$^{*}$ is is group of non-zero real number under usual multiplication

Answer 1

Since $\mathbb{R}^{*}$ is abelian, kernel of the map should contain commutator subgp of $S_3$.