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It is well-known that an element of commutator subgroup may not be a commutator.

Now there is a character-theory technique to check if an element of a group is a commutator.

An element $g\in G$ is a commutator in $G$ if and only if $$\sum_{\chi} \frac{\chi(g)}{\chi(1)}\neq 0.$$

The examples of finite groups in which some element of $G'$ is not a commutator are not so easy to produce (many examples are constructed by some ad-hoc method). For such groups, it may be difficult to quickly obtain its character table with minimum tools of character theory. The above theorem appears in chapter 3 of character theory by Isaacs, and so I will consider only these chapters.

Question: Is there simple example of a finite group, for which, we can easily determine its character table, and illustrate that some element of its commutator subgroup is not a commutator.

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  • $\begingroup$ What does "we can easily determine its character table" mean? Who are we? I usually use a computer to determine a character table, which is very quick and easy. I would guess that there are examples in which an expert in character theory (such as Isaacs) could compute the character table by hand, but I a doubtful whether you or I could do that easily. $\endgroup$ – Derek Holt Apr 28 '17 at 10:09
  • $\begingroup$ I was thinking about such possible comments. But, if to determine character table we can use computer, then to check whether an element of $G'$ is commutator we can use computer; so all the things only with black-board are pointless for purpose. I think, such questions may have been visited by some people and it is not always shared everywhere; it is difficult to get this in any book. But, MathStack could be a good platform, and I posted it here. $\endgroup$ – Beginner Apr 28 '17 at 10:12
  • $\begingroup$ It is not hard to construct examples in which it is easy to prove that not every element of the commutator subgroup is a commutator (I can explain how if you like), but such examples are generally larger than the smallest example, which has order $96$. $\endgroup$ – Derek Holt Apr 28 '17 at 10:16
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You might want to read I. M. Isaacs: Commutators and the Commutator Subgroup. The American Mathematical Monthly. Vol. 84, No. 9 (Nov., 1977), pp. 720-722

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