# 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

## 2 Answers

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.

• This works well. I chose not to argue by contradiction, and that lead to a longer argument. Nov 4, 2022 at 11:59

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
• A variation is to consider a prime for which the Sylow subgroup is not normal (which exists as the group is not abelian while all its Sylow subgroups are). This leads to the same argument in the last part, but it doesn't identify the prime as precisely. Nov 6, 2022 at 8:02