# Authorea test workspace.

## Theorems

• Let $$H$$ be a subgroup of a group missing$$G$$. 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 $$G$$.

\label{LeftCosetsDisjoint}

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.∎∎

\label{SizeOfLeftCoset}

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 $$xh_{j}$$ for $$j=1,2,\ldots,m$$. Suppose that $$j$$ and $$k$$ are integers between $$1$$ and $$m$$ for which $$xh_{j}=xh_{k}$$. Then $$h_{j}=x^{-1}(xh_{j})=x^{-1}(xh_{k})=h_{k}$$, and thus $$j=k$$, since $$h_{1},h_{2},\ldots,h_{m}$$ are distinct. It follows that the elements $$xh_{1},xh_{2},\ldots,xh_{m}$$ are distinct. We conclude that the subgroup $$H$$ and the left coset $$xH$$ both have $$m$$ elements, as required.∎∎

(Lagrange’s Theorem)\label{Lagrange} 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 \ref{LeftCosetsDisjoint}). Therefore every element of $$G$$ belongs to exactly one left coset of $$H$$. Moreover each left coset of $$H$$ contains $$|H|$$ elements (Lemma \ref{SizeOfLeftCoset}). Therefore $$|G|=n|H|$$, where $$n$$ is the number of left cosets of $$H$$ in $$G$$. The result follows.∎∎

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