## 4.7 Cosets and Lagrange’s Theorem

**Definition 4.13 **Let \((G, *)\) be a group, *H* a subgroup of *G*, and \(g \in G\).

- The left coset of
*H*by*g*is \(gH := \{ g*h : h \in H\}\). - The right coset of
*H*by*g*is \(Hg := \{ h*g : h \in H\}\).

We write \(G:H\) for the set of left cosets of *H* by elements of *G* so \(G:H = \{ gH : g \in G\}\), and \(|G:H|\) for its size. Similarly \(H:G\)
is the set of right cosets of *H* by elements of *G*.

Let’s look at some example cosets.

**Example 4.8 **

- The 3-cycle \((1,2,3) \in S_4\) has order 3, so \(H=\langle (1,2,3) \rangle\) is equal to \(\{ e, (1,2,3), (1,2,3)^2 = (1,3,2)\}\). Then \[\begin{align*} (1,2)H &= \{ (1,2), (1,2)(1,2,3), (1,2)(1,3,2) \} \\ &= \{ (1,2), (2,3), (1,3) \} \\ H(1,2) &= \{ (1,2), (1,2,3)(1,2), (1,3,2)(1,2) \} \\ &= \{ (1,2), (1,3), (2,3) \}. \end{align*}\] So in this case, \((1,2)H = H(1,2)\).
- The 2-cycle \((1,2) \in S_3\) has order 2, so \(H := \langle (1,2) \rangle\) is equal to \(\{ e, (1,2) \}\). Then \[\begin{align*} (2,3)H &= \{ (2,3), (2,3)(1,2)\} = \{ (2,3),(1,3,2)\} \\ H(2,3)&= \{ (2,3), (1,2)(2,3) \} = \{ (2,3), (1,2,3) \} \end{align*}\] So in this case \((2,3)H \neq H(2,3)\).
- We know that \(2\ZZ = \{ 2 z : z \in \ZZ\}\), the set of all even integers, is a subgroup of \((\ZZ,+)\). Then the coset \(1+2\ZZ = \{ 1+ 2z : z \in \ZZ\}\) is the set of all odd integers. More generally if \(n \in \ZZ\) then \(n\ZZ\) is a subgroup of \(\ZZ,+\) and \(1 + n\ZZ\) is the set of all integers congruent to 1 mod
*n*, \(2+n\ZZ\) is the set of all integers congruent to 2 mod*n*, and so on.

Consider the cosets \(0+3\ZZ=3\ZZ, 1+3\ZZ, 2+3\ZZ\) of the subgroup \(3\ZZ\) of
\((\ZZ,+)\). Each element of \(\ZZ\) belongs to
exactly one of them, for any integer is congruent to exactly one
of 0, 1, or 2 mod 3. So we have
\[\begin{equation*}
\ZZ = 3\ZZ \sqcup (1+3\ZZ) \sqcup (2+3\ZZ)
\end{equation*}\]
(the symbol \(\sqcup\) means that this is a disjoint union). Furthermore,
\(3n \mapsto i+3n\) is a bijection between \(3\ZZ\) and \(i+3\ZZ\), so all
three cosets have the same size. So the cosets of the subgroup \(3\ZZ\)
split \(\ZZ\) into three disjoint pieces, each of the same size. If \(\ZZ\)
were a finite set this would imply that its size was three times
that of the subgroup \(3\ZZ\). When we
prove Lagrange’s theorem, which says that if *G* is finite and *H* is a
subgroup then the order of *H* divides that of *G*, our strategy will be
to prove that you get exactly this kind of decomposition of *G* into a
disjoint union of cosets of *H*.

**Example 4.9**The

*3*-cycle \((1,2,3) \in S_3\) has order 3, so \(H= \langle (1,2,3) \rangle\) is equal to \(\{ e, (1,2,3), (1,3,2) \}\). Then \[\begin{align*} (1,2)H& = \{(1,2), (1,2)(1,2,3), (1,2)(1,3,2) \} \\ &= \{ (1,2), (2,3), (1,3) \} \\ (2,3)H &= \{ (2,3), (2,3)(1,2,3), (2,3)(1,3,2) \} \\ &= \{ (2,3), (1,3), (1,2) \} \end{align*}\] So \((1,2)H = (2,3)H\).

This last example is important, because it tells us we can have \(g_1H = g_2 H\) even if \(g_1 \neq g_2\). The next result tells us exactly when this happens.

From now on I’ll be lazy and write \(gh\) instead of \(g*h\).

**Lemma 4.8**(the

**coset equality lemma**) Let

*G*be a group and

*H*a subgroup of

*G*. Then \(gH=kH\) if and only if \(g^{-1}k \in H\), and \(Hg = Hk\) if and only if \(kg^{-1} \in H\).

*Proof. * We prove only the statement about left cosets, the proof for right cosets being analogous.

If \(gH=kH\) then \(k \in gH\), so \(k = gh\) for some \(h\in H\), so multiplying on the left by the inverse of *g* we get \(g^{-1}k \in H\).

**Corollary 4.1**If \(gH\cap kH \neq \emptyset\) then \(gH = kH\)

*Proof.*Let \(x \in gH \cap kH\). Then \(x=gh_1 = kh_2\) for some \(h_1, h_2 \in H\). Rearranging gives \(g^{-1}k = h_1h_2^{-1}\) which is an element of

*H*because

*H*is a subgroup. By the coset equality lemma, \(gH=kH\).

Of course, there’s a version of this for right cosets as well: $

Each coset of a subgroup *H* has the same size as *H*.

**Lemma 4.9**\(|gH| = |H| = |Hg|\).

*Proof. * We will prove the first of these equalities, the second being
similar. The proof is by writing down a bijection between *H*
and \(gH\). Define \(\alpha : H \to gH\) by \(\alpha(h) = g*h\).

\(\alpha\) is one-to-one: for if \(\alpha(h_1)=\alpha(h_2)\) then \(g*h_1=g*h_2\), so multiplying on the left by \(g^{-1}\) we get that \(h_1=h_2\).

\(\alpha\) is onto: for any element of \(gH\) equals \(g*h\) for some \(h \in H\), and \(\alpha(h)=g*h\), so \(g*h\) is in the image of \(\alpha\).

Thus \(\alpha\) is a bijection, and \(|gH| = |H|\).**Theorem 4.2 **(**Lagrange’s Theorem**) Let *G* be a finite group and let *H* be a subgroup. Then \(|G| = |G:H||H| = |H:G||H|\).

In particular, Lagrange’s theorem tells you that the order of a subgroup of a finite group divides the order of the group. A group of order 4 cannot possibly have a subgroup of order 3.

*Proof. * Let the distinct left cosets of *H* in *G* be \(g_1H,\ldots,g_mH\), so
\(m=|G:H|\). We will show that every element of *G* belongs to exactly one of these cosets. For any \(g\in G\) we have \(g \in gH\), so every element of \(G\) belongs to at least one of these cosets. Corollary 4.1 shows that the cosets \(g_1H, \ldots, g_mH\) don’t have any elements in common, so

**Corollary 4.2**Let

*G*be a finite group and \(g \in G\). Then the order of

*g*divides the order of

*G*.

*Proof.*The order of

*g*equals the order of the cyclic subgroup \(\langle g \rangle \leq G\) by Lemma 4.7. This divides \(|G|\) by Lagrange’s theorem.

**Corollary 4.3**Let

*G*be a group whose order is a prime number

*p*. Then

*G*is cyclic.

*Proof.*Let \(e \neq g \in G\). Consider the subgroup \(\langle g \rangle \leq G\). Its size divides \(|G|=p\) by Lagrange’s theorem, so it is

*1*or

*p*, but it is larger than one as it contains

*e*and

*g*so \(|\langle g \rangle|=p\). So \(\langle g \rangle = G\) and

*G*is cyclic.

**Corollary 4.4**(Fermat’s Little Theorem). Let \(a \in \ZZ\) and let

*p*be a prime number. Then \(a^p \equiv a \mod p\).

*Proof.*If

*a*is divisible by

*p*this is easy, so let’s suppose

*a*is not divisible by

*p*. Since \(\ZZ_p^\times, \times\) is a group of order \(p-1\), the order of \([a]_p\) divides \(p-1\) so \([a]_p^{p-1}=[1]_p\), that is, \([a^{p-1}]_p=[1]_p\). Therefore \(a^{p-1} \equiv 1 \mod p\), and so \(a^p \equiv a \mod p\).