Finite group is abelian if the representatives of its conjugacy classes commute

Let $G$ be a finite group and let $g_1 , g_2 ,...,g_r$ be the representatives of its conjugacy classes. If $g_i g_k=g_k g_i$ for every $i,k \in$ {$1,2,...,r$}, then prove that $G$ is abelian.

My original trial was to prove this in arbitrary subgroup of $S_n$ and using Cayley's theorem we can prove this easily, and the reason is we can take advantage of the fact that if $\tau , \sigma \in S_n$, $$\sigma = (a_{11} ... a_{1 n_{1}})(a_{21} ... a_{2 n_{2}}) ... (a_{r1} ... a_{r n_{r}})$$

then $$\tau \sigma \tau ^{-1} = (\tau(a_{11}) ... \tau(a_{1 n_{1}}))(\tau(a_{21}) ... \tau(a_{2 n_{2}})) ... (\tau(a_{r1}) ... \tau(a_{r n_{r}}))$$

But this didn't give me any valuable results, so any hints which can be useful ?

I found a proof of this exercise here, but I want to construct my own proof.

Any hints ?

Solution 1:

Some ideas, assuming that what you meant is that given any complete set of representatives of the different conjugation clases then the set is commutative

Let $\,g,h\in G\;$ and let $\,[[a]]\,,\,[[b]]\,$ be the corresponding conjugacy classes each of these two elements belong to. First, note that

$$[a,b]=1\implies [a,b^z]=1\;,\;\;\forall\,z\in G\;,\;\;\text{because}\;[[b]]=[b^z]]\;,\;\text{so}:$$

$$g=a^x=x^{-1}ax\;,\;\;h=b^y=y^{-1}by\;,\;\;\text{for some}\;\;x,y,\in G\;,\;\;\text{but then:}$$

$$[g,h]=[a^x,b^y]=[a,b^{yx^{-1}}]^x=1^x=1\;\;\;\text{and we're done}$$

Solution 2:

To address the question asked:

I don't believe your idea will work without some pretty heavy extra work. The problem is you choose an arbitrary permutation representation (or even the regular representation) rather than choosing a group action that is relevant to the question.

If you only consider $S_n$-conjugacy (cycle type), then you run into the following problematic example (the regular representation of the dihedral group of order 8):

• (),
• (1,2,3,8)(4,5,6,7), (1,8,3,2)(4,7,6,5),
• (1,3)(2,8)(4,6)(5,7), (1,4)(2,7)(3,6)(5,8), (1,5)(2,4)(3,7)(6,8), (1,6)(2,5)(3,4)(7,8), (1,7)(2,6)(3,5)(4,8)

If we only consider $S_n$-conjugacy, then we can choose commuting representatives: an element of order 4, its square, and the identity. These commute, but they only generate a cyclic subgroup of index 2 rather than the whole group.

Hence you have to be careful which “$\tau$” you allow, which makes your proof a lot more complicated.

The linked solution looks at a group action that is relevant: the group acting on itself by conjugation.

The linked solution however can be rephrased: Suppose for some $x \in G$, you can find conjugacy class representatives $g_i$ that each commute with $x$. Then for an arbitrary $y \in G$, there is some $g$ such that $y^g = g_i$ commutes with $x$, so $y^g \in C_G(x)$ and $y \in C_G(x)^{g^{-1}}$. In particular, $G$ is the union of the conjugates of $C_G(x)$, but a finite group is not the union of conjugates of a proper subgroup, so $G=C_G(x)$ and $x$ is in the center. Choosing $x=g_i$, you get $g_i \in Z(G)$ and $g_i$ is the only member of its conjugacy class. Since this can be done for all of the $g_i$, we get that every element of $G$ is in the center, and $G$ is abelian.