Let $G$ an commutative group of order $2n$ where $n$ is odd prime. Suppose $G$ has an element of order $n$. Show that $G$ is cyclic.
I know that every group of even order has an element of order $2$. I don't know whether it is required or not.
One can apply the the concept of upto the chapter Cyclic group and First counting principle, $o(HK) = o(H)o(K)/o(HK)$.
Since, up to this problem, the book that I am following did not introduce Lagrange's theorem, Cauchy's Theorem for Abelian Groups, Sylow's Theorem for Abelian Groups, Cayley's Theorem, Quotient group, Normal Subgroup, Isomorphism etc.
Please help me to solve the problem using the elementary tools under the above restriction.
Best Answer
Yes, you will need to use the fact that $G$ has an element of order $2$. Let $x \in G$ have order $2$, and $y \in G$ have order $n$. Here is a hint. Using the fact that $G$ is abelian and the fact that $\gcd(2, n) = 1$, show that the order of $xy$ is $2n$, whence $G$ must be cyclic.
Here is a proof that any group of even order must have an element of order $2$. Let $|G| = 2n$, and let $\Omega = \{x \in G \mid x \neq x^{-1}\} \subseteq G$, i.e. $\Omega$ is the set of all elements of $G$ which do not have order $1$ or $2$. Note that if $x \in \Omega$, then $x^{-1} \in \Omega$, and $x \neq x^{-1}$ by the hypothesis that $x \in \Omega$. Hence, $\Omega$ has even order, so $G \setminus \Omega$ has even order. Since $e \in G \setminus \Omega$, $G \setminus \Omega$ has at least order $2$, whence $G$ must have an element of order $2$.