# Abstract Nonsense

## 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 $gH$left 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

December 29, 2010 -

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