Prove that $S_3/C_3$ is a (quotient) group

Question

Consider $S_3$ to be the symmetries of a triangle, and let $C_3$ be a subgroup that cycles the three corners, so generated by: $(1 \ 2 \ 3 )$.

Using Lagrange's theorem, compute $k = |S_3/C_3|$. Then, write down $\sigma_1, \sigma_2 \dots \sigma_k \in S_3$ such that: $$ S_3/C_3 =\{\sigma_1 C_3 , \sigma_2 C_3 \dots \sigma_k C_3 \}$$ Finally, prove this is a group.

Lagrange's theorem tells us that: $$ |S_3/C_3|=|S_3|/|C_3|= 6/3 =2$$ So we know $k=2$, there are 2 equivalence classes in $S_3/C_3$. We know that reflections have order two, hence a reflection would be a good candidate to divide the set into two equivalence classes. We pick $\sigma_1=(1 \ 2), \sigma_2= \sigma^2 = id \in S_3$ such that: $$ S_3/C_3 =\{(1 \ 2) C_3, C_3 \}= \{C_3, (1 \ 2) C_3\} $$

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We now need to prove this is a group. First of all closure does not make sense to me, how do we know $C_3 (1 \ 2) C_3$ does not take us out of this set? Associativity would be inherited from associativity of permutations. I suppose the identity would be $C_3$, but how to make this argument? I'm also kind of stuck on inverses, each has to be its own inverse but I struggle to express myself formally.

Answer 1

I would use a different approach in this problem:

Since you have found that $\bigg|\dfrac{S_3}{C_3}\bigg|=2$ we can conclude that $C_3 \lhd S_3$ so $\dfrac{S_3}{C_3}$ is a group.

Answer 2

You can also use the fact that $C_3\cong A_3$, the alternating group, which is normal in $S_3$