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.


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}


\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)



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

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


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


  1. […] stated in our last post on Lagrange’s theorem that it was powerful. Indeed, it itself has wonderful, wonderful properties. But, we can use it to […]

    Pingback by Review of Group Theory: Consequences of Lagrange’s Theorem « Abstract Nonsense | December 29, 2010 | Reply

  2. […] Proof: This is evident from Lagrange’s Theorem. […]

    Pingback by Review of Group Theory: Normal Subgroups and Quotient Groups (Pt. II) « Abstract Nonsense | January 1, 2011 | Reply

  3. […] the right hand side of the above a coset of […]

    Pingback by Review of Group Theory: The First Isomorphism Theorem « Abstract Nonsense | January 2, 2011 | Reply

  4. […] This follows immediately from the First Isomorphism Theorem and Lagrange’s theorem. More […]

    Pingback by Review of Group Theory: Interesting Consequence of the First Isomorphism Theorem « Abstract Nonsense | January 2, 2011 | Reply

  5. […] Let act on by multiplication on the cosets of and let latex Gto S_p$ and . From Lagrange’s Theorem we know […]

    Pingback by Review of Group Theory: Group Actions (Pt. I Definitions and a Sharpening of Cayley’s Theorem cont.) « Abstract Nonsense | January 4, 2011 | Reply

  6. […] prior posts we’ve discussed Lagrange’s theorem and some of the profound consequences it can have on […]

    Pingback by Review of Group Theory: Group Actions (Pt. II Orbits and the Orbit Decomposition Theorem) « Abstract Nonsense | January 5, 2011 | Reply

  7. […] By the above theorem we have that is non-trivial and so by Lagrange’s Theorem we have that . Assume that then and thus (from previous theorem) we may conclude that is […]

    Pingback by Review of Group Theory: Group Actions (Pt. IV Conjugation and the Class Equation Pt. II) « Abstract Nonsense | January 6, 2011 | Reply

  8. […] thus it follows that for some . But, recalling that we may then conclude that where . But, by Lagrange’s Theorem we may conclude […]

    Pingback by Review of Group Theory: Alternate Proof of the Sylow Theorems (Pt. II) « Abstract Nonsense | January 12, 2011 | Reply

  9. […] of , namely the center of the character. We also saw how a lot of information pertaining to the index of in […]

    Pingback by Representation Theory: The Index of the Center of a Character « Abstract Nonsense | March 9, 2011 | Reply

  10. […] we know that where is the conjugacy class containing and ‘s centralizer. It follows from Lagrange’s theorem that and thus is a prime power. But, by proposition # 2 this implies that can’t be […]

    Pingback by Representation Theory: Burnside’s Theorem « Abstract Nonsense | March 11, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: