# 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

Advertisements

December 29, 2010 -

## 7 Comments »

1. […] non-trivial and so by Lagrange’s Theorem we have that . Assume that then and thus (from previous theorem) we may conclude that is cyclic. But, from an earlier theorem we may then 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

2. […] we know that has a subgroup of order and since this subgroup must be cyclic (see here for proof) we know that for some and for and thus contains at least elements of order . Since the […]

Pingback by Review of Group Theory: Sylow’s Theorems « Abstract Nonsense | January 8, 2011 | Reply

3. […] and thus . Note though that this implies that and thus by our lemma we know that . But, by an earlier theorem we know […]

Pingback by Review of Group Theory: Sylow’s Theorems (Pt. II) « Abstract Nonsense | January 8, 2011 | Reply

4. […] prove this by induction. Namely fix an arbitrary prime and we induct on groups of order . If then every group is abelian and thus trivially solvable. So, assume that every group of order is prime and let be a group of […]

Pingback by Review of Group Theory: Solvable Groups « Abstract Nonsense | March 11, 2011 | Reply

5. […] First note, using the very basic corollary of Lagrange’s theorem we know that for every and so in particular and thus so that for every , in particular for […]

Pingback by Representation Theory: The Number of Self-Conjugate Irreps On a Finite Group of Odd Order « Abstract Nonsense | March 27, 2011 | Reply

6. […] that since we have that , and since , and we  have by our lemma that . Similarly, one has that (recalling that since one has that […]

Pingback by Relation Between Sylow’s Theorems and Direct Product « Abstract Nonsense | April 19, 2011 | Reply

7. […] Since we know that and thus . Let then , since is odd and we know   we know that is odd and so for some . Thus, that said if is the canonical projection onto then […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2004) « Abstract Nonsense | May 6, 2011 | Reply