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While studying abstract algebra one of the basic and important theorems taught is the Lagrange's Theorem. And often it comes with pretty few corollaries... I was wondering whether any of these corollaries is equivalent to Lagrange's Theorem. For eg. Given a finite group G which satisfies the property that any element to the power of the order of G gives the identity. Can Lagrange's Theorem be proved using this statement.....?

Bernard
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Ivin Babu
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  • See also https://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory – lhf Sep 27 '19 at 10:52

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There are several equivalent formulations of Lagrange's theorem, but the corollaries are usually not equivalent. To give an example, Lagrange's theorem states that for any finite group $G$, the order of every subgroup $H$ of $G$ divides the order of $G$. The following statement is equivalent to Lagrange:

For all subgroups $H$ of a finite group $G$ we have $|G|=(G:H)\cdot |H|$.

One of the first corollaries of Lagrange is that every group of prime order is cyclic. This is not equivalent to Lagrange.

Lagrange's theorem can also be used to show that there are infinitely many primes. However, the infinitude of primes is not equivalent to Lagrange's theorem.

Lagrange's theorem also implies Euler's theorem, i.e., that $a^{|G|}=e$ for all $a\in G$. Again it is not equivalent to Lagrange.

One has to argue here certainly more, but it should be rather clear.

Dietrich Burde
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  • I don't expect it t be, but in what precise sense Euler's theorem isn't equivalent to Lagrange's theorem? – lhf Sep 27 '19 at 10:18
  • @lhf Good question. Like the Riemann hypothesis is equivalent to Robin's inequality. So from Euler's theorem alone you cannot conclude Lagrange, without using elements of the proof for Lagrange itself. – Dietrich Burde Sep 27 '19 at 10:45