# Cauchy's vs Lagrange's theorem in Group Theory

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?

• Which one do you think gives an equivalence? I don't really get what you are asking – Václav Mordvinov May 27 '18 at 11:01
• Cauchy's theorem is not an implication of Lagrange's theorem. – timotheechalamet May 27 '18 at 11:03
• typo: $|G|=|G:H||H|$ (not $|G|=|H:G||H|$). – drhab May 27 '18 at 11:04
• @drhab thank you, corrected it! – Jhonny May 27 '18 at 16:01
• @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. – 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 $4$.