4.1: Cosets and Lagranges Theorem

xDefinition: Cosets

Let \(G\) be a group. Let \(H \le G\) and \(g \in G\). Then

a) left coset of \(H\) with representative \(g\in G\) is denoted by \(gH=\{gh|h\in H\}\).

b) right coset of \(H\) with representative \(g\in G\) is denoted by \(Hg=\{hg|h\in H\}\).

 

Note: if \(G\) is abelian then \(H\) will be abelian, therefore \(gH=Hg, \; \forall g\in G\). The right cosets and left cosets are the same. Also, since \(H\) is a group,  \(hH=H=Hh, \forall h \in H\).  

Example \(\PageIndex{1}\)

Let \(G=\mathbb{Z}_6=\{0, 1, 2, 3, 4, 5\}\) with the operation addition \((mod \,6)\).

Let \(H=\{0,3\}\) which is a subgroup of \(\mathbb{Z}_6\).  We must ensure that \(H\) is a subgroup by ensuring closure and that the identity of \(G\) is an element of \(H\) because \(H\) is finite. Notice that \(0 \in H\).

\( +(mod 6)\)	\(0\)	\(3\)
\(0\)	\(0\)	\(3\)
\(3\)	\(3\)	\(0\)

 Hence, by the Cayley table, \(H\) is a subgroup of \(G\).

Left cosets of \(H\) will be:

\(0  + H = \{0, 3\}.\)  Note: this is equal to \(H\) and also equal to \(3 + H. \)

Note: the count of cosets is based on unique sets; thus, \(0+H=3+H\) counts as one left coset. 

\(1 + H = \{1, 4\}. \) Note: this is equal to \(4 + H.\)

\(2 + H = \{2, 5\}.\) Note that this is also equal to \(5+ H.\)

  Therefore, there are \(3\) left cosets in \(H\).
 

Note, in this case, the right cosets will be the same sets as the left cosets, since \(\mathbb{Z}_6\) is commutative.

Theorem \(\PageIndex{1}\)

Let \(G\) be a group, and let \(H \le G\). Then,

 \(gH=H=Hg\) if and only if \(g\in H.\)

Proof

Left to the reader.

Example \(\PageIndex{2}\)

Let \(G=S_3=\{e, (1,2),(1,3),(2,3),(1,2,3),(3,2,1)\}\) with the operation composition.

a) Let \(H=A_3=\{e, (1,2,3),(3,2,1)\}\).

Without showing the work, \(eA_3=A_3\) and \(eA_3=(1,2,3)A_3=(3,2,1)A_3=A_3\).

Consider \((1,2)A_3=\{(1,2)e, (1,2)(1,2,3),(1,2)(3,2,1)\}\)

  \(=\{(1,2), (2,3), (1,3)\}\).

\((1,3)A_3=\{(1,3)e, (1,3)(1,2,3),(1,3)(3,2,1)\}\)

     \(=\{(1,3), (1,2), (2,3)\}\).

\((2,3)A_3=\{(2,3)e, (2,3)(1,2,3),(2,3)(3,2,1)\}\)

     \(=\{(2,3), (1,3), (1,2)\}\).

Therefore \((1,2)A_3=(1,3)A_3=(2,3)A_3\).  

Thus we have 2 left cosets of \(H\), \(A_3\) and \(\{(2,3), (1,3), (1,2)\}\).

To calculate the right cosets, we have \(A_3G\).  Applying a process similar to that to determine the left cosets, we obtain the number of right cosets, which happen to be equal to the number of left cosets.

b) Let \(K=\{e, (1,2)\}\). Then \(eK=(1,2)K=K.\) 

Consider

\((2,3)K=\{(2,3)e,(2,3)(1,2)\}=\{(2,3), (1,3,2)\}=(1,3,2)K.\)

\((1,3)K=\{(1,3)e,(1,3)(1,2)\}=\{(1,3), (1,2,3)\}=(1,2,3)K.\)

Hence, three left cosets of \(K\) in \(G\) exist.

Now,  \(Ke=K(1,2)=K.\) 

Consider, 

\(K(2,3)=\{e(2,3),(1,2)(2,3)\}=\{(2,3), (1,2,3)\}=(1,2,3)K.\)

\(K(1,3)=\{e(1,3),(1,2)(1,3)\}=\{(1,3), (1,3,2)\}=(1,3,2)K.\)

Hence, three right cosets of \(K\) in \(G\) exist.

Notice that the number of elements in the left coset is the same as in the right coset. But \(K(2,3) \ne (2,3)K.\)

Theorem \(\PageIndex{2}\)

Let \(G\) be a group. Let \(H\) be a finite subgroup of \(G\). Then \(|gH|=|H|=|Hg|\)  for all \(g \in G\). 

Proof

Let \(G\) be a group and let \(H\) be a finite subgroup of \(G\). Let \(g \in G\). Define \(\phi: H \to gH\) by \(\phi(h)=gh\), for \(h\in H\).

We shall show that \(\phi\) is well-defined bijection. Clearly, \(\phi\) is well defined. 

\(\phi\) is injective: Let \(h_1,h_2 \in H\) such that \(\phi(h_1)=\phi(h_2)\). Then \(gh_1=gh_2\). Since \(g \in G\), \(g^{-1}\left(gh_1\right)=g^{-1}\left(gh_2\)\right) \implies \left(g^{-1}g\right)h_1=\left(g^{-1}g\right)h_2 \implies h_1=h_2.\) Hence \(\phi\) is injective.

\(\phi\) is surjective:  Let \(gh \in gH\). then  \(h\in H\) such that \(\phi(h)=gh\). Thus \(\phi\) is surjective.

 Since  \(\phi\) is a well-defined bijection, \(|H|=|gH|\),  for all \(g \in G\).  Similarly we can show that \(|H|=|Hg|\) for all \(g \in G\). 

Theorem \(\PageIndex{3}\)

Let \(G\) be a group and let \(H\le G\).  For \(g_1, g_2 \in G\), the following statements are equivalent:

1. \(g_1H=g_2H\).  Does this mean \(g_1=g_2\) or just that there is a \(g_1\) and \(g_2\) that will generate \(H\)?

2. \(g_2^{-1}g_1H=H\).

3. \(Hg_1^{-1}=Hg_2^{-1}\).

4. \(g_2 \in g_1H\).

5. \(g_1H\subseteq g_2H\).

 

Theorem \(\PageIndex{4}\)

Let \(G\) be a group and let \(H\le G\). Let \(\mathcal{L}_H=\{gH \mid| g \in G \}\)  set of all left cosets of \(H\) in \(G\), and \(\mathcal{R}_H=\{Hg \mid| g \in G \}\) set of all right cosets of \(H\) in \(G\). Then \(|\mathcal{L}_H|=| \mathcal{R}_H|.\)

Proof

Let \(G\) be a group and let \(H\le G\). Define \(\phi: \mathcal{L}_H \to \mathcal{R}_H\) by \(\phi(gH)=Hg^{-1}\), for \(g\in G\).

We shall show that \(\phi\) is well-defined bijection. Clearly, \(\phi\) is well defined. 

\(\phi\) is injective: Let \(g_1,g_2 \in G\) such that \(\phi(g_1H)=\phi(g_2H). \) Thus, \(Hg_1^{-1}=Hg_2^{-1}\). By Theorem \(\PageIndex{3}\),  \(g_1H=g_2H\). Hence the result.

\(\phi\) is surjective:  Let \(Hg \in \mathcal{R}_H\). Then  \(g^{-1}\in G\) ,\( g^{-1}H\mathcal{L}_H\) and  \(\phi(g^{-1}H)=Hg\). Thus \(\phi\) is surjective.

 Since  \(\phi\) is a well-defined bijection, \(|\mathcal{L}_H|=|\mathcal{R}_H|.\)

 

Example \(\PageIndex{3}\)

Let \(G\) be a group and let \(H \le G\).   Define a relation on \(G\) by \(g_1 \sim g_2\) is the same as \(g_1H=g_2H\), for \(g_1, g_2 \in G\).  Then we can show that \(\sim\) is an equivalence relation (i.e., it is reflexive, symmetric, and transitive).  Then \(G= \underset{g \in G}{\cup} gH\), either \(g_1H=g_2H\) or \(g_1H \cap g_2H= \{\}\).

Theorem \(\PageIndex{5}\)

Let \(G\) be a group and let \(H\le G\). Then, the set of all left cosets of \(H\) in \(G\) is a partition of \(G\). That is \(G= \underset{g \in G}{\cup} gH\) and

 either \(g_1H=g_2H\) or \(g_1H \cap g_2H= \{\}\), for \(g_1, g_2 \in G\).

Proof

Let \(G\) be a group and let \(H\le G\). Every \(g \in G) is an element of \(gH\), so clearly \(G \subset \underset{g \in G}{\cup} gH\) and the other inclusion is obvious.  Thus, \(G= \underset{g \in G}{\cup} gH\).

Let \(g_1, g_2 \in G.\) such that \(g_1H \cap g_2H\ne \{\}\). Thus there exist \( y \in g_1H \cap g_2H\). Hence \( y \in g_1H \) and \( y \in g_2H\). Thus \( y = g_1h_1 = g_2h_2\) for some \(h_1, h_2 \in H\). Which implies \(g_1 = \left(g_2h_2 \right)h_1^{-1} =g_2 \left(h_2h_1^{-1}\right) \in g_2H. \) By Theorem \(\PageIndex{3}\),  \(g_1H=g_2H\). Hence the result.

Definition: Index

 Let \(G\) be a group and \(H \le G\).

The index of \(H\) in \(G=[G:H]\) Note:  \([G:H]\) is an integer.

         \(=\) the number of left cosets of \(H\) in \(G.\)

         \(=\) the number of right cosets of H in \(H\) in \(G.\)

Theorem \(\PageIndex{4}\): Lagranges Theorem

Let \(G\) be a finite group and \(H \le G\). Then the order of \(H\) divides the order of \(G\) and \(|G|=[G:H]|H|\).

Note:\([G:H]=\frac{|G|}{|H|}\).

Proof

Let \(|G|=n\) and \(H \le G\) of order \(k.\) There are finitely many left cosets of \(H\) in \(G,\)  suppose \(m\) of them. Then by Theorem \(\PageIndex{5}\), \(G= g_1H\cup\cdots \cup g_mH\), where \(g_i \in G, \forall i=1, \cdots m.\) Then by Theorem \(\PageIndex{2}\), \(n=mk\). Hence the order of \(H\) divides the order of \(G\) and \(|G|=[G:H]|H|\).

Example \(\PageIndex{4}\)

Since \(|S_3|=6\) and let \(|K|=2\) then \([S_3:K]=\frac{6}{2}=3\), which is what we obtained in the previous example.

Example \(\PageIndex{5}\)

Using Lagrange's Theorem, we can prove that \(A_4\) has no subgroup of order 8. 

Since\(|A_4|=12\), by Lagrange’s theorem, the possible subgroups of \(A_4\) will have orders of 1, 2, 3, 4, 6, or 12.  Thus, there will not be a subgroup of order 8.

Lemma \(\PageIndex{1}\)

Let \(G\) be a group and \(H \le G\).

 If the index of \(H\) in \(G=[G:H] =2,\) then \(gH=Hg, \; \forall \, g \in G.\)

Proof

Add texts here. Do not delete this text first.

Caution

The converse of Lagrange's Theorem is not necessarily true. 

 Counterexample: Consider \(A_4\). \(6\) divides \(|A_4|\), but \(A_4\) has no subgroup of order \(6\).

\(A_4=\{e, (1\, 2\, 3), (1\, 3\, 2), (1\, 2\, 4), (1\, 4\, 2), (1\, 3\, 4), (1\, 4\, 3), (2\, 3\, 4), (2\, 4\, 3), (1\, 2)(3\, 4), (1\, 3)(2\, 4), (1\, 4)(2\, 3)\}\).

Answer

Assume that \(H\) is a subgroup of \(A_4\) such that \(|H|=6.\) Since \(|H|=6,\)  at least one of the \(3-\) cycle in \(A_4\) must be in \(H\).Without loss of generality, we may assume \((1\, 2\, 3) \in H\). Since \(H\) is a subgroup of \(A_4\), h must contain \((1\, 2\, 3)^{-1}=(3\, 2\, 1)\).

Since \([A_4,H]=\frac{12}{6}=2\),  \(gH=Hg, \; \forall g\in A_4\). Thus  \(gHg^{-1}\in H, \; \forall g\in A_4.\)

Choose  \(g=( 1\, 2\, 4) \in A_4\). Then \((1\, 2\, 4)(1\, 2\, 3)(1\, 2\, 4)^{-1} =(1\, 2\, 4)(1\, 2\, 3)(4\, 2\, 1)=(2\, 4\, 3)\in H\). Since  \(H\) is a subgroup of \(A_4\) , \((2\, 4\, 3)^{-1}=(3\, 4\, 2)\in H\). Now,  \((2\, 4\, 3)(1\, 2\, 3)(2\, 4\, 3)^{-1} =(2\, 4\, 3)(1\, 2\, 3)(3\, 4\, 2)=(1\, 4\, 2)\in H.\)

Thus \(H=\{e, (1\, 2\, 3), (1\, 3\, 2), (1\, 2\, 4), (1\, 4\, 2), (1\, 3\, 4), (1\, 4\, 3), (2\, 3\, 4), (2\, 4\, 3), \cdots\}\). Thus \(H\) has more than \(6\) elements. This is a contradiction.  Hence \(A_4\) has no subgroup of order \(6\).

Note

The converse of Lagrange's Theorem is true for finite cyclic groups (see Theorem 2.4.8). (2.4: Cyclic groups)

The following are some consequences of Lagrange's Theorem:

Corollary \(\PageIndex{1}\)

Let \(G\) be a finite group of order \(n\), and let \(g \in G\). Then the order of \(g\) divides \(n\).

Proof

Let \(G\) be a finite group of order \(n\), and let \(g \in G\). Suppose \(H=\langle g \rangle\). Then \(|H|=|\langle g \rangle|\). Hence by  Lagrange's Theorem \(|H| \mid n.\) Thus, the order of \(g\) divides \(n\). 

Corollary \(\PageIndex{2}\)

Let \(G\) be a finite group of order \(n\), then \(g^n=e\) for all \(g \in G.\)

Proof

Let \(G\) be a finite group of order \(n\), and let \(g \in G\). Then by Corollary \(\PageIndex{1}\),  the order of \(g\) divides \(n\). Suppose the order of \(g=k\). Then \(n=mk\) for some \(m \in \mathbb{Z}.\) Now, Consider, \(g^n=g^{mk}=\left(g^k\right)^m=e^m=e.\) Hence, \(g^n=e\) for all \(g \in G.\)

Corollary \(\PageIndex{3}\)

Every group of prime order is cyclic.

Proof

 Another proof can be found in  Theorem 2.4.9 (2.4: Cyclic groups).

 Let \(G\) be a finite group of order \(p\), where \(p\) is prime. Then there exists \(g \in  G\), not the identity \(e\).  Suppose \(H=\langle g \rangle\). Hence by  Lagrange's Theorem \(|H| \mid p.\) Since \(p\) is prime, \(|H|=1\) or \(|H|=p.\)  Since \(g \ne  e \), \(|H| \ne 1\). Hence \(|H|=p.\)  Thus \(G=H\). Therefore, \(G\) is cyclic.

Note

The proof of the Corollary \(\PageIndex{2}\) shows that if \(G\) is a finite group of prime order, then every non-identity element of \(G\) is a generator of \(G\).

Caution

\(\mathbb{Z}_6\) is cyclic but the order is not prime.

 

Corollary \(\PageIndex{3}\)

If \(K \le H \le G\) then \([G:K]=[G:H][H:K]\).

Example \(\PageIndex{6}\)

Let \(H\) and \(K\) be subgroups of a finite group \(G\), and gcd\((|H|,|K|)=1.\) Then show that \(H \cap K= \{e\}.\)

Answer

Since \(H\) and \(K\) be subgroups of a finite group \(G\), \(H \cap K \le H,\) and \(H \cap K \le H,\). By  Lagrange's Theorem \(|H \cap K| \mid |H|\) and  \(|H \cap K| \mid |K|.\) Since gcd\((|H|,|K|)=1,|H \cap K|=1.\) Hence, \(H \cap K= \{e\}.\)