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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 $M_a$.
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.
Let $q$ be the smallest prime factor of $[G:Z(G)]$. Let $a$ be an element of order $q$. As $q\nmid |Z(G)|$ we know that $a\notin Z(G)$, so the centralizer $C_G(a)$ is a proper subgroup of $G$. Clearly $a\in C_G(a)$ as well as $Z(G)\le C_G(a)$. Therefore $|C_G(a)|$ is divisible by $q\cdot |Z(G)|$. The number of conjugates of $a$ is equal to $m=[G:C_G(a)]>1$. By the minimality of $q$, all the prime factors of $m$ are $>q$. In particular, $m>q$.
As there are only $q$ elements in $\langle a\rangle$, the element $a$ has a conjugate $b\notin\langle a\rangle$. Obviously $b$ also has order $q$. Equally obviously $b$ and $a$ cannot commute for then they would generate a subgroup of order $q^2$, forbidden by Lagrange.
