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

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: $$S_3/C_3 =\{(1 \ 2) C_3, C_3 \}= \{C_3, (1 \ 2) C_3\}$$

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.

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