Section 10 Cosets and The Theorem of Lagrange

10.2 Definition of Cosets

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=102&selection=491,1,561,0&color=note|p.97]]

Let H be a subgroup of a group G. The subset $aH=\{ah|h\in H\}$ of G is the left coset of H containing a, while the subset $Ha=\{ha|h\in H\}$ is the** right coset** of H containing a.

10.3 Example

Example

$3\mathbb{Z}=\{\cdots ,-3,0,3,\cdots \}$

The left cosets of $3\mathbb{Z}$ of $\mathbb{Z}$ are:

• $1+3\mathbb{Z}$
• $2+3\mathbb{Z}$ The right cosets of $3\mathbb{Z}$ of $\mathbb{Z}$ are:
• $3\mathbb{Z}+1$
• $3\mathbb{Z}+2$

NOTE

For a subgroup H of an abelian group G, the partition of G into left cosets of H and the partition into right cosets are the same. Trivial

10.4 Example

partition of ${\mathbb{Z}}_6$ into cosets of the subgroup $H=\{0,3\}$ the cosets are $\{0,3\}\{1,4\}\{2,5\}$ We see the partition of the group again forms a group. This group is called as Factor Group

The Theorem of Lagrange

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=105&selection=241,0,254,1&color=note|p.100]]

Every coset (left or right) of a subgroup $H$ of a group $G$ has the same number of elements as $H$. This means the order of $H$ divides the order of $G$

Proof:

The proof relies on the concept of cosets. A left coset of $H$ in $G$ is a set of the form $gH=\{gh\mid h\in H\}$ for some $g\in G$. Similarly, a right coset is $Hg=\{hg\mid h\in H\}$. We will focus on left cosets, but the argument is analogous for right cosets.

1. Cosets partition G: The left cosets of H partition G. This means that every element of G belongs to exactly one left coset of H. To see this:

• Every element $g\in G$ belongs to the coset $gH$ (since $g=ge$ and $e\in H$).
• If two cosets $g_{1}H$ and $g_{2}H$ intersect (i.e., have a common element), then they are equal. Suppose $g_1{h}_1=g_2{h}_2$​ for some $h_1,h_2\in H$. Then $g_1=g_2{h}_2{h}_1-1$​. Since $H$ is a subgroup, $h_2{h}_1-1\in H$. Let $h_3=h_2{h}_1-1$​. Then $g_1=g_2{h}_3$​, and therefore $g_{1}H=g_2{h}_{3}H=g_{2}H$. Thus if two left cosets intersect they are equal.

2. All cosets have the same size: Every left coset $gH$ has the same cardinality as H. This can be shown by defining a bijection $f:H\to gH$ by $f(h)=gh$. This function is clearly surjective, and it’s injective because if $g{h}_1=g{h}_2$​, then $h_1=h_2$ by the cancellation law in $G$.

3. Order of G is the sum of coset sizes: Since the cosets partition $G$ and all have size $|H|$, the order of $G$ is the number of cosets multiplied by the size of each coset. Let n be the number of distinct left cosets of $H$ in $G$. Then $|G|=n|H|$.

10.11 Theorem

NOTE

Every group of prime order is cyclic

a non-identity element g: $H=⟨g⟩\in G$

$$⟨g⟩=gk\mid k\in \mathbb{Z},$$

which has size $ord(g)$. Lagrange says $ord(g)$ divides $|G|=p$. Since$g\ne e,ord(g)>1$The only divisors of the prime p are 1 and p. Therefore $ord(g)=p$

10.13 Definition

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=106&selection=91,2,120,1&color=yellow|p.101]]

The number of left cosets of H in G is the index (G : H ) of H in G.

10.14 Theorem

[[Math/ Books/A First Course in Abstract Algebra.pdf#page=106&selection=189,0,261,1&color=note|p.101]]

Suppose H and K are subgroups of a group G such that K ≤ H ≤ G, and suppose (H : K ) and (G : H ) are both finite. Then (G : K ) is finite, and (G : K ) = (G : H )(H : K )

$$(G:H)=|G|/|H|$$ $$(G:H)(H:K)=|G|/|H|\ast |H|/|K|=|G|/|K|=(G:K)$$