Prove that every group of prime order is cyclic..
let $G$ be a group and let $<g>$ $\in G$.
$<g>=<1,g,g^2,g^3,….>. $ is a subgroup of G.
Since the order is prime $n= 1 or P $,
Since g$\neq$1 , $n\geq2$ or $n=p$.
By Lagrange theorem
$[G]=[G:H]*|H|$ implies…….
I can't make ends meet to the last part..$\frac{[G]}{<g>}$
Can anyone guide me to the end of this problem?
Best Answer
Let $G$ be a group of prime order say $p$.
Take a non identity element $g\in G$ and consider group generated by $g$ i.e., $\big<g\big>$.
As the group is finite $|\big<g\big>|=n$ for some positive integer $n$.
Now, As $\big<g\big>\leq G$, we thank Lagrange theorem and conlcude that $|\big<g\big>|$ divides $|G|$
i.e., $n$ divides $p$.
But $p$ being prime has no proper factors which gives only possibilities of $n$ to be $1$ or $p$.
As $g$ is a non identity element $n\neq 1$ which implies $n=p$
we have $\big<g\big>\leq G$ and $|\big<g\big>|=|G|$ which implies $\big<g\big>=G$ i.e., $G$ is cyclic.