# Let G be a non abelian group with a square-free order. Prove that there are two elements $a,b\in G$ such that $ab\neq ba$ and $ord(a)=ord(b)$.

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 $$M_a$$.

• I too would argue by contradiction, but I'd look at $H$ a maximal abelian/cyclic subgroup. Since it would have to commute with all of its conjugates it would be normal. Then with any $x\notin H$ we'd have $G=\langle H,x\rangle$ and we'd be able to reduce to the case where $G=\langle y,x\rangle$ with $x,y$ of prime order. Nov 4, 2022 at 10:59

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.
• Thinking about groups like $S_3$, $C_7\rtimes C_3$, $D_n$ (with $n$ odd and squarefree) and direct product of those with abelian groups of coprime order etc pinpointed the smallest prime outside the center as a candidate. The possibility of all the conjugates of $a$ being its powers was resolved that way as well. Nov 6, 2022 at 6:41