Show that every proper subgroup of $S_3$ is cyclic.

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Show that every proper subgroup of $S_3$ is cyclic. So I approached it like this, $$|S_3|=6$$ So divisors of 6 are 2 and 3 (excluding 1 and 6, because improper subgroups). Both 2 and 3 are prime and any group of prime order is cyclic. Is that a correct approach?

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That's a correct approach, yes. I think that a more natural approach (but that's a matter of taste) would be to make the list of all proper subgroups of $S_3$ (there are only four such subgroups) and to check that each of them if cyclic.