Let $G$ be a non abelian finite group of order $p_1p_2p_3\cdot…\cdot p_n$ where $n\in \mathbb{N}, n\geq2$ and $p_i$ is a prime for every $i$. Prove that there are two elements $a,b\in G$ such that $ab\neq ba$ and $ord(a)=ord(b)$.
My idea for this problem was to firstly assume the contrary, and then to look at all the sets like $M_a$ that contains all elements from $G$ of order $a$ because a element from $M_a$ commutes with any other member of $M_a$.
Best Answer
Continuing your reasoning.
Then the subgroup $H_a=\langle M_a\rangle$ is a normal abelian subgroup of the group $G$. This is true for any element $a\in G$ of order $p_i$. If the order of $a$ is $p_i$, then it follows from the condition that $|H_a|=p_i$. This means that every Sylow subgroup of $G$ is normal and abelian, but then $G$ is also abelian. Contradiction.