# Lagrange's Theorem

**Lagrange's theorem** is a statement in group theory which can be viewed as an extension of the number theoretical result of Euler's theorem. It is an important lemma for proving more complicated results in group theory.

## Preliminary

For any group \(G,\) \(G\) itself and \(\{e\}\) are subgroups of \(G.\) Most groups have other, non-trivial subgroups. The following are some examples of subgroups, that correspond to the example list of groups from the post Group Theory.

1) \(\mathbb{Z} \), the set of integers.

2) \( \mathbb{R}^+ \), the set of positive real numbers.

3) \( S_{n-1} \), the set of functions which also satisfy \( f(n) = n \).

4) \( m \mathbb{Z}_n \), the subset of multiples of \(m\). Note that this is strictly smaller than \(\mathbb{Z}_n\) if \(\gcd(m,n)\neq 1 .\)

5) \((\mathbb{Z}_n ^\times )^2 \), the set of squares modulo \(n\).

A general example of a subgroup is obtained by picking an element \(g \in G \) and taking all powers of \(g\). This is known as the **subgroup generated by \(g\)**, and is denoted by

\[ < g > := \{ \ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots \}. \]

Proposition:The

order of the subgroup\( < g > \) is the smallest positive \(m\) for which \(g^m = e \). If such an \(m\) does not exist, then the order is infinite.

As such, we define the **order of element \(g\)** to be the smallest positive \(m\) for which \( g^m = e \), and write \( o(g) = m \).

Let \(H\) be a subgroup of the group \(G\), and let \(g \in G \). The set \( g H = \{ gh : h \in H \} \) is called a (left) **coset** of \(H\) in \(G.\) In other words, the coset \(gH\) is the image (range) of the function \( \phi_g : H \rightarrow G ,\) where \( \phi_g (h) = gh \). Because this function is one-to-one (by the cancellation property!), \(\lvert gH \rvert = \lvert H \rvert.\) Note also that \(g=ge\in gH.\)

The most important property of the cosets is that \(g_1H\) and \(g_2H\) either coincide or do not intersect (see Worked Example 2 below). This implies the following classical theorem.

## Statement of the Theorem

Lagrange's Theorem.If \(G\) is a finite group and \( H \) is a subgroup of \(G\), then \( \lvert H \rvert \) divides \( \lvert G \rvert \).

As an immediate corollary, we get that if \(G\) is a finite group and \(g\in G \), then \( o(g) \) divides \( \lvert G \rvert \). In particular, \( g^{ \lvert G \rvert} = e \) for all \( g \in G \).

## Worked Examples

Prove the proposition.

Suppose \(m\) is finite. We claim that \( < g > = \{e, g, g^2, \ldots, g^{m-1} \} \) and that all these \(m\) elements are distinct. For the first statement, note that for any \(n \in \mathbb{Z}\), we can write \(n = qm+r,\) where \( 0 \leq r \leq m-1 \). Hence \(g^n = (g^m)^q(g^r) = e(g^r) = g^r\). To prove the second statement, suppose \(g^i = g^j\) for \( 0 \leq i \leq j \leq m-1 \), then \(g^{j−i} = e\), which contradicts the minimality of \(m\). \(_\square\)

Let \( g, g' \in G \) and \( H \subseteq G \) be a subgroup.

(a) If \( g^{-1} g' \in H \), then \( gH = g'H\).

(b) If \(g^{-1} g' \not \in H \), then \( gH \cap g'H = \emptyset\).

For the first statement, any element of \(g'H\) can be written as \(g'h\), for some \(h\in H \). But \( g' h = g(g^{-1} g') h \in gH \) since \(g^{-1} g' \in H \). This tells us that \( g'H \subseteq gH \).

Conversely, any element of \( gH \) can be written as \(gh, h \in H \) and \( gh = g'(g'^{−1}g)h\). But \( g'^{−1}g = (g^{−1}g')^{−1}\) lies in \(H\) since \(H\) is a subgroup of \(G\). Hence the result follows and \(gH = g′H\).For the second statement, suppose \( x \in gH \cap g'H \). Then it can be written as \( x = gh = g'h'\) for some \( h, h' \in H \). Thus \(g^{−1}g' = hh'^{-1} \in H,\) which contradicts the fact that \( g^{−1}g' \not \in H \). \(_\square\)

Prove Lagrange's theorem.

The above shows that the cosets partition the entire group \(G\) into mutually disjoint subsets, all of which have \( \lvert H \rvert \) elements. Hence, Lagrange's theorem follows. \(_\square\)

Show that if \(|G|\) is prime, then \(G\) is cyclic.

Taking any element \(x \in G \setminus \{e_G\}\), by Lagrange's theorem, the order of \(x\) must be either 1 or \(|G|\). Since the only element of \(G\) with order 1 is \(e_G\), \(x\) has order \(|G|\). Thus, \(G\) is generated by \(\{x\}\), so \(G\) is cyclic. \(_\square\)

Show that if the orders of groups \(G\) and \(H\) are relatively prime, then their intersection contains only the identity.

The first thing to note is that \(G \cap H\) is a subgroup of both \(G\) and \(H\). Then, by Lagrange's theorem, \(|G \cap H|\) divides both \(|G|\) and \(|H|\). Since \(\text{gcd}(|G|,|H|) = 1\), we must have \(|G \cap H| = 1\). Only one subgroup of order 1 exists, which is the group consisting of only the identity, so \(G \cap H = \{e\}\). \(_\square\)

**Cite as:**Lagrange's Theorem.

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