Cosets and Lagrange’s theorem

Section 1.11 Cosets and Lagrange’s theorem

Definition 1.11.1. Group operations on subsets.

Let \(A\) be a subset of the group \(G\text{.}\) Given \(g\in G\text{,}\) the sets \(gA\text{,}\) \(Ag\text{,}\) \(gAg^{-1}\) are defined as follows:

\begin{align} A^{-1} \amp =\{a^{-1}\mid a\in G\}\tag{1.11.1}\\ gA \amp =\{ga\mid a\in A\}\tag{1.11.2}\\ Ag \amp =\{ag\mid a\in A\}\tag{1.11.3}\\ gAg^{-1} \amp =\{gag^{-1}\mid a\in A \}\text{.}\tag{1.11.4} \end{align}

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Definition 1.11.2. Coset and coset space.

Let \(H\) be a subgroup of the group \(G\text{.}\) Given \(g\in G\) the set \(gH\) defined as

\begin{equation} gH =\{gh\mid h\in H\}\tag{1.11.5} \end{equation}

is called the left \(H\)-coset in \(G\) corresponding to \(g\text{,}\) and the set \(Hg\) defined as

\begin{equation} Hg=\{hg\mid h\in H\}\tag{1.11.6} \end{equation}

is called the right \(H\)-coset in \(G\) corresponding to \(g\text{.}\) Elements of a coset are called representatives of that coset.

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The set of all left (resp. right) \(H\)-cosets is called the left (resp. right) coset space of \(H\text{,}\) and is denoted \(G/H\) (resp. \(H\backslash G\)).

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Notation 1.11.3. Cosets are left cosets by default.

Unless stated otherwise, a coset in this course will be assumed to be a left coset.

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Example 1.11.4. Cosets in \(\Z\).

Recall that any subgroup \(H\) of \(\Z\) can be described as \(H=(n)\) for a unique \(n\in \Z_{\geq 0}\text{.}\) Show that the cosets of \(H=(n)\) are precisely the congruence classes modulo \(n\text{.}\)

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Solution.

Fix \(H=(n)\text{.}\) Given \(a\in \Z\) the coset \(a+H\) is by definition

\begin{align*} a+(n) \amp =\{a+qn\mid q\in \Z\}\\ \amp = [a]_n\text{.} \end{align*}

This computation shows cosets are congruence classes and vice versa.

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Example 1.11.5. Cosets in \(\Z/12\Z\).

Let \(G=\Z/12\Z\) and let \(H=\angvec{\overline{4}}\leq G\text{.}\) Compute \(G/H\text{.}\)

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Solution.

We have \(H=\angvec{\overline{4}}=\{0,4,8\}\text{.}\) A simple computation shows

\begin{align*} \overline{0}+H \amp = H\\ \overline{1}+H \amp =\{\overline{1},\overline{5},\overline{9}\}\\ \overline{2}+H \amp =\{\overline{2},\overline{6},\overline{10}\}\\ \overline{3}+H \amp = \{\overline{3},\overline{7},\overline{11}\}\\ \overline{4}+H \amp =\{\overline{4},\overline{8},0\}=\{\overline{0},\overline{4},\overline{8}\}=H\text{.} \end{align*}

This suggests that \(G/H=\{\overline{0}+H,\overline{1}+H,\overline{2}+H,\overline{3}+H\}\text{.}\) This can be shown by proving that \(\overline{a}+H=\overline{b}+H\) if and only if \(a\equiv b\pmod 4\text{.}\) Theorem 1.11.7 below takes care of this for us.

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Example 1.11.6. Cosets in \(D_4\).

Let \(G=D_4\) and let \(H=\angvec{rs}\text{.}\) Compute \(G/H\) and \(H\backslash G\text{.}\)

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Solution.

We have \(H=\angvec{rs}=\{1, rs\}\text{.}\) We compute

\begin{align*} 1H \amp =H\\ rH= \amp \{r,r^2s\}\\ r^2H \amp =\{r^2, r^3s\}\\ r^3H \amp =\{r^3,s\}\\ sH \amp = \{s,r^3\}=r^3H \\ (rs)H \amp =\{rs, e\}=H\\ (r^2s)H \amp =\{r^2s, r\}=rH\\ (r^3s)H \amp =\{r^3s, r^2\}=r^2H \end{align*}

We conclude that \(G/H=\{H,rH,r^2H,r^3H\}\text{.}\) Let’s look at the right cosets:

\begin{align*} H1 \amp =H\\ Hr \amp =\{r,s\}\\ Hr^2 \amp =\{r^2,r^3s\}\\ Hr^3 \amp = \{r^3, r^2s\}\\ Hs \amp =\{s,r\}=Hr\\ H(rs) \amp =\{rs, e\}=H\\ H(r^2s) \amp =\{r^2s,r^3\}=Hr^3\\ H(r^3s) \amp =\{r^3s, r^2\}=r^2H\text{.} \end{align*}

We conclude that \(H\backslash G=\{H, Hr, Hr^2, Hr^3\}\text{.}\)

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Theorem 1.11.7. Cosets and coset space.

Let \(H\) be a subgroup of the group \(G\text{.}\)

The relation \(g\sim g'\) if and only if \(g'=gh\) for some \(h\in H\) is an equivalence relation and the cosets \(gH\) are the corresponding equivalence classes.
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Given \(g,g'\in G\text{,}\) the following statements are equivalent.
1. \(gH=g'H\text{.}\)
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2. \(g'\in gH\text{.}\)
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3. \(g^{-1}g'\in H\text{.}\)
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4. \(gH\cap g'H\ne\emptyset\text{.}\)
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The cosets of \(H\) form a partition of \(G\text{.}\) That is, we have

\begin{align*} G \amp =\bigcup_{g\in G} gH \end{align*}

and

\begin{equation*} gH\ne g'H\iff gH\cap g'H=\emptyset\text{.} \end{equation*}

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Proof.

This is another example of a theorem where we get a lot of mileage out of defining an equivalence relation. The relation \(g\sim g'\) if and only if \(g'\in gH\) is easily seen to be an equivalence relation:

\begin{align*} g=ge \amp \implies g\in gH\\ \amp \implies g\sim g\\ g \sim g' \amp \implies g'\in gH \\ \amp \implies g'=gh\\ \amp \implies g=g'h^{-1}\in g'H\\ \amp \implies g'\sim g\\ g\sim g' \text{ and } g'\sim g''\amp \implies g'=gh \text{ and } g''=g'h' \\ \amp \implies g''=g'h'=g(hh')\in gH\\ \amp \implies g\sim g''\text{.} \end{align*}

Since the equivalence classes of \(\sim\) are precisely the cosets \(gH\) by definition, statement (3) and most of the equivalent statements in (2) follow immediately. The exception is the statement that \(gH=g'H\) if and only if \(g^{-1}g'\in H\text{,}\) which follows using equivalent formulation (b), since

\begin{align*} gH=gH' \amp \iff g'\in gH\\ \amp \iff g'=gh \text{ for some } h\in H\\ \amp \iff g^{-1}g'=h \text{ for some } h\in H\\ \amp \iff g^{-1}g'\in H\text{.} \end{align*}

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Theorem 1.11.8. Lagrange’s theorem.

Let \(H\) be a subgroup of the group \(G\text{.}\)

We have

\begin{align} \abs{G} \amp = \abs{H}\abs{G/H}\text{.}\tag{1.11.7} \end{align}

When \(G\) is infinite, this can be interpreted as saying \(H\) is infinite or \(G/H\) is infinite.

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If \(\abs{G}< \infty\text{,}\) then we have

\begin{align} \abs{G/H}\amp = \frac{\abs{G}}{\abs{H}} \text{.}\tag{1.11.8} \end{align}

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Proof.

Statement (2) follows directly from (1). Statement (1) follows from Theorem 1.11.7 and a counting argument. First observe that for all \(g\in G\) we have \(\abs{H}=\abs{gH}\) as witnessed by the map

\begin{align*} H \amp \rightarrow gH\\ h \amp \longmapsto gh\text{,} \end{align*}

which has inverse map

\begin{align*} gH \amp \mapsto H\\ g' \amp \longmapsto g^{-1}g' \text{.} \end{align*}

Next, let \((g_{\alpha}H)_{\alpha\in I}\) be the family of distinct cosets, so that we have

\begin{align*} \abs{G/H} \amp =\abs{I}\\ G\amp=\bigcup_{\alpha\in I} g_{\alpha}H \end{align*}

and \(g_\alpha H\cap g_\beta H=\emptyset\) for all \(\alpha\ne \beta\text{.}\) If \(I\) and \(\abs{H}\) are both finite, then the sum rule for disjoint sets yields

\begin{align*} \abs{G} \amp =\sum_{\alpha\in I}\abs{g_{\alpha}H}\\ \amp = \abs{I}\cdot \abs{H} (\abs{g_{\alpha}H}=\abs{H})\\ \amp =\abs{G/H}\abs{H}\text{.} \end{align*}

This both proves the result in the case where \(G\) is finite, and shows that if \(G\) is infinite, than either \(H\) or \(G/H\) is infinite.

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Definition 1.11.9. Index of subgroup.

Given a subgroup \(H\) of the group \(G\text{,}\) its index \(\abs{G\colon H}\) is defined as

\begin{equation*} \abs{G\colon H}=\abs{G/H}\text{.} \end{equation*}

In other words, \(G\colon H\) is the cardinality of the coset space \(G/H\text{.}\)

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Remark 1.11.10. Lagrange’s theorem.

Using the index notation, we can re-write Lagrange’s theorem as

\begin{equation*} \abs{G}=\abs{H}\abs{G\colon H}\text{.} \end{equation*}

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Corollary 1.11.11. Cardinality of subgroups.

Assume \(G\) is a finite group.

For any subgroup \(H\leq G\text{,}\) we have

\begin{equation} \abs{H}\mid \abs{G}\text{.}\tag{1.11.9} \end{equation}

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For any \(g\in G\) we have

\begin{equation} \ord g\mid \abs{G}\text{.}\tag{1.11.10} \end{equation}

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Corollary 1.11.12. Groups of prime cardinality.

If \(\abs{G}=p\) for a prime integer \(p\text{,}\) then \(G\) is cyclic. It follows that

\begin{equation*} G\cong \Z/p\Z\text{.} \end{equation*}

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