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This is probably a basic question, but how would I compute $S_3/(123)$? I've figured out that this group must be abelian, as $(123)$ is the commutator subgroup and hence the quotient must be abelian, but I'm not sure where this gets me. We also know this quotient group must be order 2.

Any help appreciated!

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    There are not many groups with two elements... – 2017-02-28
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    Please note that a better notation would be $S_3/\langle (123) \rangle$. – 2017-02-28

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The factor cosets are $\{(),(123),(132)\}$ and $\{(12),(13),(23)\}$. There is only one group with $2$ elements, $Z_2$, and there isn't enough space for anything except abelian behaviour.