# Abstract Nonsense

## Review of Group Theory: Consequences of Lagrange’s Theorem

Point of post: In this post we give a few  interesting theorems which are near-corollaries of Lagrange’s Theorem.
Motivation

We 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 prove other, beautiful theorems. That is what this post is about.

This first one, although very simple shall be one of the most useful in our next post.

Theorem: Let $G$ be a finite group and $g\in G$. Then, $g^{|G|}=e$.

Proof: We know from Lagrange’s theorem that $|g|$ divides $|G|$ and thus there exists some $k\in\mathbb{N}$ such that $|g|k=|G|$. Thus,

$\text{ }$

$g^{|G|}=g^{|g|k}=\left(g^{|g|}\right)^k=e^k=e$

$\blacksquare$

Theorem: Let $G$ be a non-trivial group. Then, $G$ has no non-trivial proper subgroups if and only if $G$ is finite and of order $p$ where $p$ is prime. Moreover, every group of order $p$ is prime is cyclic.

Proof: First assume that $G$ has no non-trivial proper subgroups. We claim that $G$ is cyclic. To do this we choose $g\in G$ with $g\ne e$.We note then that $\{e\}<\left\langle g\right\rangle\leqslant G$ and thus by assumption we must have that $G=\left\langle g\right\rangle$. We recall though this implies that $G\cong\mathbb{Z}$ or $G\cong\mathbb{Z}_n$ for some $n\in\mathbb{N}$. Clearly we can’t have that $G\cong \mathbb{Z}$ otherwise there would exist some isomorphism $\theta:\mathbb{Z}\to G$ and $\{e\}<\theta\left(2\mathbb{Z}\right) which is a contradiction. Thus, $G$ is finite. Now let $n$ be such that $G\cong\mathbb{Z}_n$ ($n=|G|$). But, we know that every cyclic group has a subgroup of order $d$ where $d$ divides the order of the ambient group. In particular for every $d$ such that $d\mid n$ $G$ has a subgroup of order $d$. It follows by assumption that $d\mid n$ implies that $d=1,n$ and thus $n$ is prime.

Conversely, if $|G$|=p is prime then every subgroup must have order which divides $p$ and thus the only subgroups are of order $1,p$ from where it follows that $G$ has no non-trivial proper subgroups. But, from the first sentence of the previous previous paragraph we may also conclude that $G$ is cyclic.   $\blacksquare$

Corollary: Let $G$ be a group with $|G|=pq$ with $p,q$ prime. Then, every proper subgroup of $G$ is cyclic.

Another interesting theorem which one can use Lagrange’s theorem to prove is the following

Theorem: Let $G$ be a finite group and $H,K\leqslant G$. Then,

$\text{ }$

$\displaystyle \#\left(HK\right)=\frac{|H||K|}{|H\cap K|}$

$\text{ }$

where $HK$ is the product set defined in our last post.

Proof: Note first that

$\text{ }$

\displaystyle \begin{aligned}\bigcup_{h\in H}hK &= \left\{c:c\in hK\text{ for some }h\in H\right\}\\ &= \left\{hk:h\in H\text{ and }k\in K\right\}\\ &= HK\end{aligned}

$\text{ }$

but since $\#\left(hK\right)=\#(K)$ for each $h\in H$  it suffices to the number of elements of $G/K$ of the form $hK,\text{ }h\in H$. To do this we note that $h_1K=h_2K$ if and only if $h_1^{-1}h_2\in K$ which is true if and only if $h_1^{-1}h_2\in H\cap K$ which is true if and only if $h_1(H\cap K)=h_2(H\cap K)$. Thus,  it follows that the distinct number of elements of $G/K$ of the form $hK,\text{ }h\in H$ is equal to $\left(H:H\cap K\right)$. But, by Lagrange’s theorem we may then conclude that

$\text{ }$

\displaystyle \begin{aligned}\#\left(HK\right) &= \#\left(\bigcup_{h\in H}hK\right)\\ &=|K| \left(H:H\cap K\right)\\ &= \frac{|H||K|}{\left|H\cap K\right|}\end{aligned}

$\text{ }$

$\blacksquare$

Along similar lines

Theorem: Let $G$ be a finite group and $H,K\leqslant G$ be such that $\left(|H|,|K|\right)=1$. Then, $H\cap K$ is trivial.

Proof: We merely note that

$\text{ }$

$|H|=\left|H\cap K\right|\left(H:H\cap K\right)$

$\text{ }$

and

$\text{ }$

$|K|=\left|H\cap K\right|\left(K:H\cap K\right)$

$\text{ }$

so that $\left|H\cap K\right|$ divides both $|H|$ and $|K|$. It follows that $\left|H\cap K\right|=1$ which implies that $H\cap K$ is trivial. $\blacksquare$

Lastly

Theorem: Let $G$ be a finite group and $H\leqslant G$ be such that $\left(G:H\right)=p$ where $p$ is prime. Then, if $H\leqslant K\leqslant G$ then $K=H$ or $K=G$.

Proof: We note by $\mathbf{(4)}$ of Lagrange’s theorem that

$\text{ }$

$\left(G:H\right)=\left(G:K\right)\left(K:H\right)$

$\text{ }$

from where it follows that either $\left(G:K\right)=1$ or $\left(K:H\right)=1$. But, noting that for any subgroup $S$ of $G$ we have that $eS=S\in G/S$ it follows that if $\left(G:S\right)=1$ then $G=S$. From this the conclusion follows. $\blacksquare$

References:

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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