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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 ?

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  • $\begingroup$ @amWhy ,I have edited the question to apply this tip. Is it ok now or does it need more editing ? i will be more careful to apply this tip in my next questions :) $\endgroup$ – Fawzy Hegab Jul 8 '13 at 15:27
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    $\begingroup$ Looks good, @MathsLover! (We'll work on punctuation, next...but the post looks fine!) $\endgroup$ – Namaste Jul 8 '13 at 15:29
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    $\begingroup$ This follows from the fact that representatives of the conjugacy classes generate the group, which has been discussed and proved in two of the posts that are coming up in the "related" list. $\endgroup$ – Derek Holt Jul 8 '13 at 18:55
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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.

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    $\begingroup$ I've 2 questions , 1) How can we know that such $x$ exist ? can we suppose $x = 1$ ? , 2) You said that those representative of $D_8$ generate cyclic subgroup of order 4 " clear" , Doesn't this contradict the fact that the representatives of conjugacy classes generate the whole of the group ? may clarify this point more plz ? . I use the notation $D_{2n}$ for Diherdrial group of order $2n$ . $\endgroup$ – Fawzy Hegab Jul 9 '13 at 3:49
  • $\begingroup$ Do you mean that x commute with all of the representatives of conjugacy classes or some of them ? $\endgroup$ – Fawzy Hegab Jul 9 '13 at 3:53
  • $\begingroup$ Ops ! , i have understood your point which i asked about on question 1 , no need to clarify it, just i need the second question and third one to be answered . $\endgroup$ – Fawzy Hegab Jul 9 '13 at 3:59
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    $\begingroup$ (2) $\langle z,z^2,1 \langle = \langle z\rangle$ is cyclic with the same order as $z$, and $z$ was chosen to have order 4. $\{z,z^2,1\}$ are not representatives of the $G$-conjugacy classes of $G$. They are representatives of the $S_n$-conjugacy classes of $G$. (3) I mean that $x$ commutes with each $g_i$. If $G$ has 13 conjugacy classes, then I assume $x$ commutes with 13 elements, one from each conjugacy class. $\endgroup$ – Jack Schmidt Jul 9 '13 at 12:45
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    $\begingroup$ Yes. The $S_n$ conjugacy class of $g \in G$ is $\{ h \in G : h=g^\sigma \text{ for some } \sigma \in S_n \}$. $\endgroup$ – Jack Schmidt Jul 10 '13 at 2:07
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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

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

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

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

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  • $\begingroup$ Hmmm...my assumption may be too restrictive since I didn't use finiteness of $\,G\,$ . $\endgroup$ – DonAntonio Jul 8 '13 at 16:35
  • $\begingroup$ What if $g$ and $h$ are conjugate? you only get that $[a,a]=[b,b]=[b^z,b^z]=1$ but this is true in all groups. $\endgroup$ – Jack Schmidt Jul 8 '13 at 18:47
  • $\begingroup$ @JackSchmidt , I've added comments on your answer , but i couldn't mention your name in the comments and I don't know the reason! $\endgroup$ – Fawzy Hegab Jul 9 '13 at 7:33
  • $\begingroup$ @DonAntonio ,Is $[a,b]$ the commutator of $a$ and $b$ ? $\endgroup$ – Fawzy Hegab Jul 9 '13 at 7:48

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