# Lagrange's Theorem

**Lemma:** Let \(H\) be a subgroup of \(G\). Let \(r, s \in G\).
Then \(H r = H s\) if and only if \(r s^{-1}\in H\). Otherwise
\(H r, H s\) have no element in common. Similarly, \(r H = s H\) if and only
if \(s^{-1} r \in H\), otherwise \(r H, s H\) have no element in common.

**Proof:** If \(r s^{-1} = h \in H\), then
\(H = H h = (H r) s^{-1}\). Multiplying both sides on the right by \(s\) gives
\(H r = H s\). Conversely, if \(H r = H s\), then since \(r \in H r\) (because
\(1 \in H\)) we have \(r = h' s\) for some \(h' \in H\). Multiplying on the right
by \(s^{-1}\) shows that \(r s^{-1}\in H\).

Now suppose \(H r, H s\) have some element in common, that is \(h_1 r = h_2 s\) for some \(h_1,h_2 \in H\). This implies \(r s^{-1} = h_1^{-1} h_2 \in H\), thus \(H r = H s\) by above.

**Lagrange’s Theorem:** If \(H\) is a subgroup of \(G\), then
\(|G| = n|H|\) for some positive integer \(n\). This is called the
*index* of \(H\) in \(G\). Furthermore, there exist
\(g_1,...,g_n\) such that \(G = H r_1 \cup ... \cup H r_n\) and similarly
with the left-hand cosets relative to \(H\).

**Proof:** Take any \(r_1 \in G\). Note
\(|H r_1 | = |H|\). If \(H r_1 \ne G\) then take any
\(r_2 \in G \setminus H r_1\). By the lemma, \(H r_1, H r_2\) are disjoint
so we have \(|H r_1 \cup H r_2| = 2|H|\). By continuing in this fashion,
after \(n\) steps for some positive integer \(n\),
we will eventually have accounted for all of the elements of \(G\).
We will have \(|G| = n|H|\) and \(G = H r_1 \cup ... \cup H r_n\).

**Corollary:** Let \(G\) be a group and \(g \in G\). Then the order of \(g\)
divides \(|G|\).

**Corollary:** Let \(G\) be a group of prime order. Then \(G\) has no subgroups
and hence is cyclic.

*blynn@cs.stanford.edu*💡