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 $Hx = \{ hx : h \in H \}$ {::comment} note: the start at end disables numbering

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.