be defined as a left coset and right coset of \(H\) in \(G\text{,}\) respectively.
Exercise3.2.Abelian Cosets.
If \(G\) is an abelian then \(gH=Hg\) for all \(g\in G\text{.}\)
Proposition3.3.Basic Coset Properties.
Let \(H\leq G\) and \(x,y\in G\)
\(\displaystyle x\in xH\)
\(xH=H\) if and only if \(x\in H\)
\(\displaystyle (xy)H=x(yH)\)
Lemma3.4.Coset Equivalencies.
Let \(H \leq G\text{.}\) The following facts about left cosets are equivalent for \(x,y \in G\text{:}\)
\(x\) and \(y\) belong to the same left coset of \(H\) in \(G\text{,}\)
\(x = yh\) for some \(h \in H\text{,}\)
\(y = xh\) for some \(h \in H\text{,}\)
\(y^{-1}x \in H\text{,}\)
\(x^{-1}y \in H\text{,}\)
\(xH = yH\text{.}\)
Proof.
\((1) \Rightarrow (2):\) if \(x\) and \(y\) belong to the same left coset \(gH\) of \(H\) in \(G\) then \(x=gh'\) and \(y=gh''\) for some \(h',h''\in H\text{,}\) so \(g=y(h'')^{-1}\) and therefore \(x=y(h'')^{-1}h'=yh\) where \(h=(h'')^{-1}h'\in H\text{.}\)
\((2) \iff (3):\)\(x=yh\) for some \(h \in H\)\(\iff\)\(y = xh^{-1}\) and \(h^{-1} \in H\text{.}\)
\((2) \iff (4):\)\(x=yh\) for some \(h \in H\)\(\iff\)\(y^{-1} x=h\in H\text{.}\)
\((4) \iff (5):\)\(y^{-1} x\in H \iff (y^{-1} x)^{-1}\in H \iff x^{-1}y\in H\text{.}\)
\((2) \Rightarrow (6):\) Suppose \(x = yh\) for some \(h \in H\text{,}\) then by \(2. \Rightarrow 3.\) we also have \(y = xh''\) for some \(h'' \in H\text{.}\) Then we have
\((6) \Rightarrow (1):\) Since \(e_G=e_H\in H\text{,}\) we have \(x=xe_G\in xH\) and \(y=ye_G\in yH\text{.}\) If \(xH=yH\) then, \(x\) and \(y\) belong to the same left coset.
Theorem3.5.
For any \(H\leq G\) and any \(g\in G\text{,}\) the set of (left) cosets induces an equivalence relation on \(G\text{,}\) where two elements are related if they belong to the same left coset of \(H\) in \(G\text{.}\)
Corollary3.6.Cosets Partition a Group.
For \(H \leq G\text{,}\) the collection of left cosets of \(H\) in \(G\) form a partition of \(G\text{,}\) and similarly for the collection of right cosets. That is,
for all \(x,y \in G\text{,}\) either \(xH = yH\) or \(xH \cap yH = \emptyset\)
\(\bigcup_{x\in G} xH=G\text{,}\)
and similarly for right cosets. Moreover all left and right cosets have the same cardinality: \(|xH| = |Hx|=|H|\) for any \(x\in G\text{.}\)
Proof.
Let me prove the assertions for right cosets. Every element \(g\) of \(G\) belongs to at least one right coset, since \(g \in Hg\) (since \(e \in H\)). We need to show any two cosets are either identical or disjoint: if \(Hx\) and \(Hy\) share an element, then it follows from \((1)\Rightarrow (6)\) of Lemma 3.4 that \(Hx=Hy\text{.}\) This proves that the right cosets partition \(G\text{.}\) To see that all right cosets have the same cardinality as \(H\text{,}\) define a function
\begin{equation*}
\rho: H \to Hg
\end{equation*}
by \(\rho(h) = hg\text{.}\) We see \(\rho\) is onto and if \(\rho(h) = \rho(h')\) then \(hg = h'g\) and hence \(h = h'\text{,}\) so that \(\rho\) is also one-to-one.
Theorem3.7.Lagrange’s Theorem.
If \(G\) is a finite group and \(H \leq G\text{,}\) then
\begin{equation*}
\begin{aligned}
|G| & = |H| \cdot (\text{the number of left cosets of $H$ in $G$}) \\
& = |H| \cdot (\text{the number of right cosets of $H$ in $G$}). \\
\end{aligned}
\end{equation*}
In particular, \(|H|\) divides \(|G|\text{.}\)
Corollary3.8.Equal Number of Left and Right Cosets.
The number of left cosets of \(H\) in \(G\) is equal to the number of right cosets of \(H\) in \(G\text{.}\)
Definition3.9.Index.
In finite groups, the common number of left or right cosets of a subgroup \(H\) in a group \(G\) is denoted as \([G:H]\) and is called the index of \(H\) in \(G\text{.}\)
Example3.10.Cosets in \(D_{2n}\).
For \(G =D_{2n}\) and \(H = \langle s \rangle = \{e,s\}\text{,}\) the left cosets of \(H\) in \(G\) are
Note that these lists are not the same, but they do have the same length. We have \(|G| = 2n\text{,}\)\(|H| = 2\) and \([G:H] = n\text{.}\)
Now that we have proved Lagrange’s Theorem, a host of other results now become available to us.
Corollary3.11.Corollaries to Lagrange’s Theorem.
Let \(G\) be a finite group.
If \(x\in G\text{,}\) then \(|x|\) divides \(|G|\text{.}\)
\(\displaystyle [G:H]=\frac{|G|}{|H|}\)
If \(|G|=n\) then \(g^n=e\) for all \(g\in G\)
The notion of index is most useful when it is finite, but note that this does not require that \(G\) to be finite.
Example3.12.Finite Index in Infinite Group.
Let \(G = \mathbb{Z}\) and \(H = 7\mathbb{Z}\text{.}\) Then \([G:H] = 7\) since the cosets are \(H, 1 + H, 2 + H, \dots, 6 + H\text{.}\) (Since \(G\) is abelian, left and right cosets are automatically the same by Exercise 3.2)
Exercise3.13.Number of Left and Right Cosets (Ininite Version).
Show that even if \(G\) is not finite the number of left and right cosets of a subgroup \(H\leq G\) is still the same.
Hint.
Consider the map \(gH\mapsto Hg^{-1}\) and show it’s a bijection. Why is the inverse needed?
This seems like it might be important.
Theorem3.14.The Index Tower.
Let \(G\) be a group (possibly infinite) and \(H \leq K\) subgroups of \(G\text{.}\) Suppose \([G : H]\) is finite. Then \([G : H] = [G : K][K : H]\text{.}\)
Let \(G\) be a group of composite order. Then \(G\) contains a non-trivial proper subgroup.
Exercise3.16.Fermat’s Little Theorem.
Prove Fermat’s Little Theorem: for every integer \(a\) and every prime \(p\text{,}\)\(a^p\mod p=a\mod p\text{.}\)
If you can believe it, there’s actually another important corollary to Lagrange’s Theorem, but we’ll cover it when its more relevant. Wouldn’t want to spoil all the fun at once, right? 1
To spoil all the fun at once, see: Theorem 6.3, part (1).
Summary
For any \(H\leq G\) and any \(g\in G\text{,}\) a left and right Cosets of \(H\) in \(G\) is defined to be
Lagrange’s Theorem states that the order of any subgroup divides the order of the group, and thus that the order of an element must also divide the order of the group. This is one of the seminal results in group theory.
