Does Lagrange's Theorem lead to the result that if |G| = 20, for example, then I can expect to find subgroups of orders 1, 2, 4, 5, 10 only or just that those are some expected subgroup orders?
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Setting aside the issue of whether the converse is true (though it's not) - it would be much more difficult to prove, since you would have to somehow construct these subgroups only by knowing their order. It's not really clear how you would go about doing this - so we shouldn't expect the proof of Lagrange's Theorem to help us, at least not without a lot more work. – Jair Taylor May 12 '17 at 07:34
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The converse actually does hold for abelian groups. But this requires the structure theorem for abelian groups. – Jair Taylor May 12 '17 at 07:37
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See this question for more on the converse of Lagarnge's Theorem. – Dietrich Burde May 12 '17 at 08:07
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If $|G| = 20 $, it means that if $H$ is a subgroup of $G$ then it is necessary that $|H| \mid |G|$. However, note that Lagrange's theorem doesn't at all guarantee the existence of subgroups with orders that are divisors of 20. That is, Lagrange's theorem doesn't guarantee the existence of a single non-trivial subgroup. Therefore Lagrange's theorem doesn't assure us of the existence of subgroups with orders 2,4,5 and 10. But it does guarantee that if we happen to find any subgroup of $G$, then it will necessarily be a divisor of 20. Remember to not confuse Lagrange's theorem with its possible converse, which is false.

erdoswiles
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You say $G$ may not have a single non-trivial subgroup. That's wrong. If Lagrange's theorem allows more than the trivial subgroups then there are more. For example there is a subgroup for each prime $p$ dividing the group order. But this isn't a consequence of Lagrange's theorem but rather the statement of Cauchy's theorem. – principal-ideal-domain May 12 '17 at 08:26
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@principal-ideal-domain I meant that in the context of what the Lagrange's theorem implies. I have made it clearer. Thanks for pointing that out. – erdoswiles May 12 '17 at 08:49