# Finite and Infinite order

• Apr 6th 2010, 02:26 AM
student17989
Finite and Infinite order
an element a of a group is the smallest positive integer m such that a^m = e (where e denotes the identity element of the group, and a^m denotes the product of m copies of a). If no such m exists, we say that a has infinite order. All elements of finite groups have finite order.

can some1 give me an easy examples using the above facts please? thanks
• Apr 6th 2010, 08:51 AM
Drexel28
Quote:

Originally Posted by student17989
an element a of a group is the smallest positive integer m such that a^m = e (where e denotes the identity element of the group, and a^m denotes the product of m copies of a). If no such m exists, we say that a has infinite order. All elements of finite groups have finite order.

can some1 give me an easy examples using the above facts please? thanks

Here are two groups. One that has infinite order and one that doesn't. While the one that doesn't may seem overcomplicated there is a reassuringly easy way to prove that it's infinite.

Finite: $S_n=\left\{\sigma:\{1,\cdots,n\}\to\{1,\cdots n\}\mid \sigma\text{ is bijective}\right\}$ with the binary operation of function composition. To prove that each element has finite order we use the following theorem:

Theorem: Let $G$ be a finite group, then for each $g\in G$ we have that $|g|<\infty$.

Proof: Suppose not. Then, $\left\{e,g,g^2,g^3,\cdots\right\}$ must all be distinct, for to suppose that $g^{k}=g^{\ell}\implies g^{|k-\ell|}=e$ contradicting the assumption that $|g|=\infty$. Thus, since $G$ is closed under multiplication it follows that $\{e,g,g^2,g^3,\cdots\}\subseteq G$ and so $|G|=\infty$. Contradiction.

Infinite: Let $\Omega$ be an infinite set and $S_\Omega$ be defined as above. Then, $|S_\Omega|=\infty$. To see this suppose not. Then, $|S_\Omega|=k\in\mathbb{N}$. Select elements $\omega_1,\cdots,\omega_k\in S_\Omega$ and consider $K=\left\{\sigma\in S_n:\sigma(m)=m,\text{ }\forall m\notin\{\omega_1,\cdots,\omega_k\}\right\}$. Clearly $K\leqslant S_\Omega$ and $\theta:S_k\hookrightarrow S_\Omega$ given by the obvious mapping is an embedding and $\theta(S_k)=K$. Thus, $|K|=k!$. But, by our assumption $|S_\Omega|=k<\infty$ and so Lagrange's theorem implies that $k!\mid k$. Contradiction.

Does that answer your question? There are infinite groups whose elements all have finite order though. For example $\bigoplus_{n=1}^\infty Z_n$ where each $Z_n$ is a copy of $\mathbb{Z}_2$