I have just started studying group theory, and we have naturally looked at Lagrange's theorem and stated Cauchy's as a collorary.

Lagrange: If $ H \subset G$ we must have $ |G| = |G:H| |H| $. Or shortly, the order of every subgroup must divide the order of the parent group. However, as far as I understand this doesn't say that if |G| = 0 (mod n), there must be subgroups of order n.

Cauchy: If the order of a group factorises $ |H| = p_1*p_2*p_3...$ there must be elements of order $p_1,p_2...$.

How comes one of them gives a mere can while the other one gives a must? I have tried looking up separate proofs to Cauchy, but they seem to surpass my group theory knowledge. In lectures we proved Lagrange by proving that every element is in one, and only one, coset. How can we easily prove Cauchy's theorem?

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    $\begingroup$ Which one do you think gives an equivalence? I don't really get what you are asking $\endgroup$ – Václav Mordvinov May 27 '18 at 11:01
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    $\begingroup$ Cauchy's theorem is not an implication of Lagrange's theorem. $\endgroup$ – timotheechalamet May 27 '18 at 11:03
  • $\begingroup$ typo: $|G|=|G:H||H|$ (not $|G|=|H:G||H|$). $\endgroup$ – drhab May 27 '18 at 11:04
  • $\begingroup$ @drhab thank you, corrected it! $\endgroup$ – Jhonny May 27 '18 at 16:01
  • $\begingroup$ @Anwi after a bit of googling I think I have realised this too, thank you! I am now just wondering if there is some accessible proof of it available. $\endgroup$ – Jhonny May 27 '18 at 16:02

The Cauchy Theorem is only true for primes. In fact the group $\mathbb{Z}_2 \times \mathbb{Z}_2$ has order $4$, but no element of order $4$.

The Lagrange Theorem tells you that a subgroup of certain order exists, then it's order must divide the order of the group. The Cauchy Theorem is some sort of a reverse implication of the Lagrange Thereom, as it works only for prime divisors of the order of the group.

  • $\begingroup$ Thank you for the clarification about the primes! Do you have any recommendation about where I can find an accessible proof of Cauchy's theorem? $\endgroup$ – Jhonny May 27 '18 at 16:03
  • $\begingroup$ Have a look at the note of K. Conrad. Moreover, this site has many proofs and further details, e.g., here, or here. $\endgroup$ – Dietrich Burde May 27 '18 at 16:07
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    $\begingroup$ @Jhonny I'm pretty sure you can find a solid proof about Cauchy Theorem on the internet. Nevertheless I would recommend reading Fraleigh's "First course in Abstract Algebra". It is suitable for beginners and also easy to understand $\endgroup$ – Stefan4024 May 27 '18 at 16:07

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