# 4.1: Cosets and Lagranges Theorem

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##### Definition: 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 assure that $$H$$ is, in fact a subgroup by ensuring there is closure, an identity exists, each element has inverse and operation on the set is associative.  In this case clearly yes.

Left cosets of $$H$$ will be:

0 (from G) + H = {0, 3}.  Note: this is equal to H and also equal to 3 + H.  Note: the count of

cosets are based on unique sets, thus 0+H=3+H counts as 1 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.  Thus $$\mathbb{Z}_6$$ is commutative.

##### Example $$\PageIndex{2}$$

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

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 coset of $$H$$, $$A_3$$ and $$\{(2,3), (1,3), (1,2)\}$$.

To calculate the right cosets, we have $$A_3G$$.  Applying the process similar to that to determine the left cosets, we obtain the right cosets which happen to be equal to the left cosets.

##### Theorem $$\PageIndex{1}$$

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 g1 and g2 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. $$g_1H\subseteq g_2H$$.

##### 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 $$g_1H=g_2H$$.  Then we can show that $$\sim$$ is an equivalence relation (ie, it is reflexive, symmetric, and transitive).  Then $$G= \underset{g \in G}{\cup} gH$$ means either $$g_1H=g_2H$$ or $$g_1H \cap g_2H= \{\}$$.

##### 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 G.

##### Theorem $$\PageIndex{2}$$: Lagranges Theorem

Let $$G$$ be a finite group and $$H \le G$$.

Then $$|G|=[G:H]|H|$$. Note:  The order of $$H$$ divides the order of $$G$$ and $$[G:H]=\frac{|G|}{|H|}$$.

##### 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.

##### Caution

The converse of Lagrange's Theorem is not true.

Consider $$A_4$$. $$6$$ divides $$|A_4|$$, but $$A_4$$ has no subgroup of order $$6$$.

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

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 $$6$$.

##### Corollary $$\PageIndex{1}$$

Every group of prime order is cyclic.

##### Theorem $$\PageIndex{3}$$

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

##### Example $$\PageIndex{5}$$

From this theorem, we can prove that $$A_4$$ has no subgroup of order 8.  $$|A_4|=12$$. From 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.

This page titled 4.1: Cosets and Lagranges Theorem is shared under a not declared license and was authored, remixed, and/or curated by Pamini Thangarajah.