Prove that 2 cyclic subgroups of same order are either equal or contain $e$ only as intersection?

Question

Can someone give me a proof that $2$ cyclic subgroups of same order in a group are either equal or their intersection is $e$. I think Lagrange's theorem is required for it. Also suggest some visual interpretation of this proof.

Answer 1

It is not true.

Consider the Quaternion group $Q_8$.
It contains two cyclic groups of same order which are
$$\langle i \rangle=\{1,i,-1,-i\}$$ and $$\langle j \rangle=\{1,j,-1,-j\}$$ Obviously, they are not equal and also their intersection which is $\{1,-1\}$ is not trivial.

Answer 2

If the order of each of the cyclic groups is the same prime number $p$, then it is true. And yes you can use Lagrange’s Theorem to prove this.