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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2Yes, this is correct. By the way, note that the trivial subgroup is also cyclic. – Mark Oct 15 '20 at 13:22
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Yes, this is correct, but it assumes that you already know the reuslt that a group of prime order is cyclic. – Dietrich Burde Oct 15 '20 at 13:23
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Thanks to @Mark – Potato Oct 15 '20 at 13:43
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Thanks @DietrichBurde – Potato Oct 15 '20 at 13:43
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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.
José Carlos Santos
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