This is the problem in the book that I want to prove, but it doesn't seem right.

For example let's say I have a group of $9$ elements. If this group is non-cyclic then every element (except identity) has an order of $3$ (prime) (because of Lagrange's theorem). So there are actually $8$ elements of order $3$.

Theorem seems to work for cyclic groups.

Is there is something I am missing here or this only holds for cyclic groups?

  • That $8$ is a multiple of $2$? – universalset Nov 27 '13 at 14:45
  • What is the problem with $p=3$? – Calvin Lin Nov 27 '13 at 14:45
  • wow, I somehow managed to miss a word "multiple of". – Minter Nov 27 '13 at 14:48

Let $X$ be the set of elements of order $p$. The relation $$x \sim y :\Longleftrightarrow \langle x \rangle = \langle y \rangle$$ is obviously an equivalence relation and the equivalence class of $x \in X$ is $\langle x \rangle \setminus \{e\}$. Thus each equivalence class has size $p-1$, i.e. $|X|$ is a multiple of $p-1$.

hint: if the order of element $g$ is $p$, what can you say about the order of element $g^n$?

Hint: If there are no such elements, then this is easy. If there exists such an element, say $g,$ then consider how many such elements are in $\langle g\rangle,$ the cyclic subgroup generated by $g$. Given any two such cyclic subgroups, if they are distinct, then what does Lagrange tell you about how many elements must they have in common? What can you then conclude?

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