# This note has been used to help create the Lagrange's Theorem wiki

In this post, we will build up towards proving **Lagrange's Theorem**, which can be viewed as an extension of Euler's Theorem.

Let \(G\) be a group with operation \(*\) and consider an arbitrary subset \(H \subseteq G \). In general, \(H\) is not a group under \(*\). In fact, \(*\) may not even be a well-defined function on \(H \times H \rightarrow H\), since its image may lie outside \(H\). Hence, we make the following definition.

A

subgroup of \(G\)is a subset \(H \subseteq G \) such that:

(a) \(e \in H \),

(b) \( h \in H \Rightarrow h^{-1} \in H \),

(c) \(h, h' \in H \Rightarrow hh' \in H \).

Note that any such \(H\) becomes a group in its own right, with the operation \(*\) "inherited" from \(G.\)

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. Theorder 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.

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 \(G\). In particular, \( g^{ \lvert G \rvert} = e \) for all \( g \in G \).

## Worked Examples

## 1. Prove the proposition.

Solution: 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 \(m\).

## 2. 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 \( gH \cap g'H = \emptyset\).

Solution: 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 \( g^{−1}g' \not \in H \).

## 3. Prove Lagrange's Theorem

Solution: 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.

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TopNewestReally useful notes Aditya – Usama Khidir · 2 years, 1 month ago

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