Abstract Nonsense

Crushing one theorem at a time

Review of Group Theory: Cosets and Lagrange’s Theorem (Pt. II)


Point of post: This is a continuation of this post.

Cosets

If H\leqslant G and g\in G we call the set gHleft coset of H in G, or just a left coset when there is no ambiguity. Similarly we define a right coset of H and denote it Hg. We denote the set of all left cosets by G/H, put more explicitly

G/H=\left\{gH:g\in G\right\}

 

and we denote \#\left(G/H\right) by (G:H), this is called the index of G in H.

Remark: It’s clear that the number of left cosets of H is equal to the number of right cosets of H. This is because it’s easy to verify f:gH\to Hg is a bijection between the two sets.

The first thing one notices about cosets is that:

Theorem: Let G be a group and H\leqslant G. Define the relation \overset{H}{\sim} on G by


g\overset{H}{\sim}h\Leftrightarrow g^{-1}h\in H


then \overset{H}{\sim} is an equivalence relation and [g]=gH (where [g] denotes the equivalence class of g under \overset{H}{\sim}).

Proof: We first prove that \overset{H}{\sim} is an equivalence relation. To do this we note that clearly g\overset{H}{\sim}g for every g\in G since gg^{-1}=e\in H. To prove that \overset{H}{\sim} is symmetric we note that if g\overset{H}{\sim}h then g^{-1}h\in H and thus (g^{-1}h)^{-1}=h^{-1}g\in H so that h\overset{H}{\sim}g. Lastly, to prove that \overset{H}{\sim} is transitive we note that if g\overset{H}{\sim}h and h\overset{H}{\sim}k then g^{-1}h\in H and h^{-1}k\in H and thus g^{-1}k=(g^{-1}h)(h^{-1}k)\in H. From these three axioms it follows that \overset{H}{\sim} is an equivalence relation.

Next, to prove that [g]=gH it suffices to note that if k\in[g] then g^{-1}k\in H so that g^{-1}k=h for some h\in H and thus k=gh for some h\in H and so k\in gH. Conversely, if k\in gH then k=gh for some h\in H and so g^{-1}k=h\in H. The conclusion follows. \blacksquare

Corollary: Let H\leqslant G then


\displaystyle \biguplus_{gH\in G/H}gH=G


(where \uplus is meant to denote that the elements of G/H are pairwise disjoint, and they’re union is all of G. In other words, G/H forms a partition of G).

Remark: Evidently the same methodology applies to show that the set of right cosets of H partitions G.

We are now ready to prove one of the main results of basic finite group theory

Theorem(Lagrange): Let G be a finite group and K\leqslant H\leqslant G and g\in G. Then:


\displaystyle \begin{aligned}&\mathbf{(1)}\quad |G|=(G:H)|H|\\ &\mathbf{(2)}\quad |H|\textit{ divides }|G|\\ &\mathbf{(3)}\quad |g| \textit{ divides }|G|\\ &\mathbf{(4)}\quad (G:K)=(G:H)(H:K)\end{aligned}


Proof:

\mathbf{(1)}: Recall that

\displaystyle G=\biguplus_{gH\in G/H}gH

 

But, we know from the first part of Cayley’s Theorem that \theta_g (translation by g) is a bijection so that |H|=\#\left(gH\right). Thus, it follows that

\displaystyle \begin{aligned}|G| &= \#\left(\biguplus_{gh\in G/H}gH\right)\\ &= \sum_{gh\in G/H}\#\left(gH\right)\\ &= \sum_{gH\in G/H}|H|\\ &= \#\left(G/H\right)|H|\\ &= \left(G:H\right) |H|\end{aligned}

\mathbf{(2)}: This follows immediately from \mathbf{(1)}.

\mathbf{(3)}: This follows from previous theorem since \left\langle g\right\rangle\leqslant G and |g|=\left|\left\langle g\right\rangle\right|.

\mathbf{(4)}: This follows immediately from \mathbf{(1)} since

\displaystyle (G:K)=\frac{|G|}{|K|}=\frac{|G|}{|H|}\frac{|H|}{|K|}=\left(G:H\right)\left(H:K\right)

\blacksquare

Reference:

1.  Lang, Serge. Undergraduate Algebra. 3rd. ed. Springer, 2010. Print.

2. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 200


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December 29, 2010 - Posted by | Algebra, Group Theory | , ,

10 Comments »

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