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Exercise 978Algebra › GroupsL1Added 3 Nov 2026

Corollary of Lagrange's theorem

#Lagrange#order of element#corollary

Statement

Let GG be a finite group of order nn. Show that for every g∈Gg\in G, gn=eg^n=e.

Hint

The order of gg equals the order of the cyclic subgroup ⟨g⟩\langle g\rangle, which divides ∣G∣|G| by Lagrange.

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Solution

Order of gg. Let dd be the order of gg (the smallest positive integer with gd=eg^d=e). Then ⟨g⟩={e,g,…,gd−1}\langle g\rangle=\{e,g,\ldots,g^{d-1}\} has order dd.

Lagrange. By Lagrange's theorem, d=∣H∣d=|H| divides ∣G∣=n|G|=n.

Conclusion. n=kdn=kd for some kk, so gn=gkd=(gd)k=ek=eg^n=g^{kd}=(g^d)^k=e^k=e.

Corollary. In (Z/pZ)∗(\mathbb{Z}/p\mathbb{Z})^* with pp prime, for any a≢0a\not\equiv0, ap−1≡1(modp)a^{p-1}\equiv1\pmod p (Fermat's little theorem).