Group Theory

Lemma: Let $H$ be a subgroup of $G$. Let $r,s\in G$. Then $Hr=Hs$ if and only if $r{s}^{-1}\in H$. Otherwise $Hr,Hs$ have no element in common. Similarly, $rH=sH$ if and only if $s^{-1}r\in H$, otherwise $rH,sH$ have no element in common.

Proof: If $r{s}^{-1}=h\in H$, then $H=Hh=(Hr)s^{-1}$. Multiplying both sides on the right by $s$ gives $Hr=Hs$. Conversely, if $Hr=Hs$, then since $r\in Hr$ (because $1\in H$) we have $r=h\prime s$ for some $h\prime \in H$. Multiplying on the right by $s^{-1}$ shows that $r{s}^{-1}\in H$.

Now suppose $Hr,Hs$ have some element in common, that is $h_{1}r=h_{2}s$ for some $h_1,h_2\in H$. This implies $r{s}^{-1}=h_1^{-1}{h}_2\in H$, thus $Hr=Hs$ by above.

Lagrange's Theorem: If $H$ is a subgroup of $G$, then $\mid G\mid =n\mid H\mid$ for some positive integer $n$. This is called the index of $H$ in $G$. Furthermore, there exist $g_1,...,g_n$ such that $G=H{r}_1\cup ...\cup H{r}_n$ and similarly with the left-hand cosets relative to $H$.

Proof: Take any $r_1\in G$. Note $\mid H{r}_1\mid =\mid H\mid$. If $H{r}_1\ne G$ then take any $r_2\in G\setminus H{r}_1$. By the lemma, $H{r}_1,H{r}_2$ are disjoint so we have $\mid H{r}_1\cup H{r}_2\mid =2\mid H\mid$. By continuing in this fashion, after $n$ steps for some positive integer $n$, we will eventually have accounted for all of the elements of $G$. We will have $\mid G\mid =n\mid H\mid$ and $G=H{r}_1\cup ...\cup H{r}_n$.

Corollary: Let $G$ be a group and $g\in G$. Then the order of $g$ divides $\mid G\mid$.

Corollary: Let $G$ be a group of prime order. Then $G$ has no subgroups and hence is cyclic.