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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?

miguel barnes
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  • You don't know g isn't 1, and it's not necessary to. What are n and p? The order of g and of the group? You're going in reverse, Lagrange's theorem is the support for n = 1 or P. Now, if it's cyclic, what is the lattice of subgroups like? – Loki Clock Sep 25 '13 at 01:51
  • Some MathJax advice:

    < and > mean "less than" and "greater than", and produce spacing correct for that meaning only; to make angle brackets, use \langle and \rangle.

     – Zev Chonoles Sep 25 '13 at 01:56
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    Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. – Zev Chonoles Sep 25 '13 at 01:57

4 Answers4

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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.

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Ok let me help you organize the ideas. You want to show that $G$ is cyclic. Assume that order of $G$ is $p$. Then, as you said, let $g$ be a non-identity element. You found that the order of $g$ is $p$. The subgroup generated by an element has order equal the order of that element. Hence, the subgroup generated by $g$ is of order $p$, call this subgroup $H$.

We have $H\leq G$, and they both have the same size, so we get $H=G$, so $G$ is generated by $g$.

Daniel Montealegre
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Let $G$ be a group of prime order,$p$ and let $a\in G$ be a non-identity element. Now by Lagrange's theorem $\dfrac{\vert G\vert}{\vert \langle a\rangle \vert}=p or 1$. If $\dfrac{\vert G\vert}{\vert \langle a\rangle \vert}=p$, then $a=e$ which is impossible and if $\dfrac{\vert G\vert}{\vert \langle a\rangle \vert}=1$, then $G=\langle a\rangle$.

Anupam
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Let $a \in G$ be a non-identity element then by $Lagrange's$ $theorem$, $|a|$ \ $|G|$ so $|a|=p$, so the cyclic subgroup generated by $a$ is the whole $G$. Hence $G$ must be cyclic.

Mathematician
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