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Math and science::Algebra::Aluffi

Groups. Lagrange's theorem.

Lagrange's theorem

Let \( G \) be a group, and let \( H \subset G \) be a subgroup. Then:

\[ |G| = |G\!/H| \; |H| \]

There are three corollaries on the reverse. Can you remember them?

  1. About the order of \( g \) in \( G \).
  2. When \( |G| \) is a prime.
  3. Fermat's little theorem.

\( |g| \) divides \( |G| \)

For any element \( g \in G \), \( |g| \) divides \( |G| \). In particular, \( |g| \) equals the order of the subgroup generated by \( g \).

As a result, \( g^{|G|} = e_G \) for any element in a finite group \( G \).

When \( |G| \) is a prime

If \( |G| \) is a prime \( p \), then we must have \( G \cong \mathbb{Z}\!/p\mathbb{Z} \).

Fermat's little theorem

Let \( p \) be a prime, and let \( a \) be an integer. Then \( a^{p} \equiv a \mod p \)

Note on Fermat's little theorem

If one expects to see \( a^{p+1} \) instead of \( a^{p} \), then realize that if one considers the sequence \( e, a, a^2, ... \), then the element \( a^{1} \) can be throught of as having already traversed two elements, \( e \) and \( a \). So, to cover all \( p \) elements and by back to \( e \) we only need \( a^{p-1} \), then another \( a \) brings us back to \( a \).

Source

Aluffi p105