Lagrange Theorem And Cyclic Subgroups

Question

In the lecture note which I copied states that one of lagrange theorem implication is that

If $G$ is a finite group and $g\in G$ then $o(g)\mid |G|$ where $o(g)=|\langle g\rangle|$

does the equality $o(g)=|\langle g\rangle|$ holds for all finite groups?

Yes. That's the definition of $o(g)$. The divisibility result $o(g)\mid |G|$ is what the theorem is saying. — Jyrki Lahtonen, Feb 20 '18 at 08:08
@JyrkiLahtonen But that does not imply that $\langle g\rangle=G$ is that correct? — gbox, Feb 20 '18 at 08:10
Correct. It doesn't, That holds only when $G$ is cyclic, and $g$ happens to be a generator. — Jyrki Lahtonen, Feb 20 '18 at 08:11

score 1 · Answer 1 · answered Feb 20 '18 at 08:27

Critical point to understand it:

Every element of finite group has an order.here

Let $G$ be group and $a\in G$ such that $o(a)=m<\infty$. Then the following conditions on a finite group $G$ are equivalent.

$\textbf{(i)}$ $a^n=e\iff m\mid n$ where $n$ is positive integer.

$\textbf{(ii)}$ $a^i= a^j\iff i\equiv j\pmod{m}$

$\textbf{(iii)}$ Elements $e=a^0, a^1, a^2, \cdots a^{m-1}$ are different.

$\textbf{(iv)}$ $\langle a\rangle=\{e, a, a^2,\cdots, a^m-1\}$

$\textbf{(v)}$ $o(a)=\mid\langle a\rangle\mid$

It is not hard to show them.

score 1 · Answer 2 · answered Feb 20 '18 at 17:50

$o(g)=$ the smallest $n\in \mathbb{N_{>0}}$ such that $g^n=1_G$

$<g>=\{g^k:k\in \mathbb{Z}\}=\{...,g^{-2},g^{-1},1_G,g,g^2,g^3,...\} =\{g^{sn+u}:s,u\in \mathbb{Z}$ and $0\leq u\leq n-1 \}=\{g^u:u\in \mathbb{Z},0\leq u\leq n-1\}\Rightarrow |<g>|=n=o(g)$

Lagrange Theorem And Cyclic Subgroups

2 Answers2

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