AMS-$\LaTeX$ Theorem Styles

Proof supports $\LaTeX$ with [AMS] extensions, which provide various features to facilitate writing math formulas and to give the general structure and appearance of a mathematics article or book.

In particular, theorem styles including definition, theorem, lemma, proof, and a few more¹ can be declared as an environment using this template:

**Theorem**{:.label #Lagrange}
  _(Lagrange's Theorem)_
  Let $$G$$ be a finite group, and let $$H$$ be a subgroup
  of $$G$$.  Then the order of $$H$$ divides the order of $$G$$.
{:.theorem}

Various components of this template can be skipped or customized, so experiment with them. For instance, to use your own enumeration, use {:.label*} instead.

An Example

The following exposition is taken from [here]. Note the minimal changes required

Lagrange’s Theorem

Definition Let \(H\) be a subgroup of a group \(G\). A left coset of \(H\) in \(G\) is a subset of \(G\) that is of the form \(xH\), where \(x \in G\) and \(xH = \{ xh : h \in H \}\). Similarly a right coset of \(H\) in \(G\) is a subset of \(G\) that is of the form \(Hx\), where \(Hx = \{ hx : h \in H \}\)

Note that a subgroup \(H\) of a group \(G\) is itself a left coset of \(H\) in \(G\).

Lemma Let \(H\) be a subgroup of a group \(G\), and let \(x\) and \(y\) be elements of \(G\). Suppose that \(xH \cap yH\) is non-empty. Then \(xH = yH\).

Proof Let \(z\) be some element of \(xH \cap yH\). Then \(z = xa\) for some \(a \in H\), and \(z = yb\) for some \(b \in H\). If \(h\) is any element of \(H\) then \(ah \in H\) and \(a^{-1}h \in H\), since \(H\) is a subgroup of \(G\). But \(zh = x(ah)\) and \(xh = z(a^{-1}h)\) for all \(h \in H\). Therefore \(zH \subset xH\) and \(xH \subset zH\), and thus \(xH = zH\). Similarly \(yH = zH\), and thus \(xH = yH\), as required.

Lemma Let \(H\) be a finite subgroup of a group \(G\). Then each left coset of \(H\) in \(G\) has the same number of elements as \(H\).

Proof Let \(H = \{ h_1, h_2,\ldots, h_m\}\), where \(h_1, h_2,\ldots, h_m\) are distinct, and let \(x\) be an element of \(G\). Then the left coset \(xH\) consists of the elements \(x h_j\) for \(j = 1,2,\ldots,m\). Suppose that \(j\) and \(k\) are integers between \(1\) and \(m\) for which \(x h_j = x h_k\). Then \(h_j = x^{-1} (x h_j) = x^{-1} (x h_k) = h_k\), and thus \(j = k\), since \(h_1, h_2,\ldots, h_m\) are distinct. It follows that the elements \(x h_1, x h_2,\ldots, x h_m\) are distinct. We conclude that the subgroup \(H\) and the left coset \(xH\) both have \(m\) elements, as required.

Theorem (Lagrange’s Theorem) Let \(G\) be a finite group, and let \(H\) be a subgroup of \(G\). Then the order of \(H\) divides the order of \(G\).

Proof Each element \(x\) of \(G\) belongs to at least one left coset of \(H\) in \(G\) (namely the coset \(xH\)), and no element can belong to two distinct left cosets of \(H\) in \(G\) (see Lemma). Therefore every element of \(G\) belongs to exactly one left coset of \(H\). Moreover each left coset of \(H\) contains \(|H|\) elements (see Lemma). Therefore \(|G| = n |H|\), where \(n\) is the number of left cosets of \(H\) in \(G\). The result follows.