Why is $S_3 / A_3$ cyclic group of order $2$?

Question

Consider symmetric group $S_3$ and alternating group $A_3$.

I wonder why quotient group $S_3 / A_3$ is cyclic group of order $2$.

I know $S_3 / A_3=\{ \sigma \circ A_3 \mid \sigma \in S_3 \}$ but I don't know why this is cyclic group of order $2$.

Thank you for your help.

Answer 1

Because $A_3\le S_3$ and their orders are $3!/2$ and $3!$, respectively; more specifically,

$$|S_3|=[S_3:A_3]|A_3|,$$

while $[S_3:A_3]=|S_3/A_3|.$

Also, the only group of order two is $\Bbb Z_2$ up to isomorphism, so it must be cyclic.