Consider that a group $A$ acts by automorphism on a finite group $G$. If this action is coprime, i.e. $\gcd(|A|,|G|)=1,$ can we affirm that this action is fixed point free, i.e. $C_G(A)=1$?

I tried to think about the order of the automorphism and the order of na element of $G$, but it doesn’t work.

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    $\begingroup$ What does "coprime" mean here? Does it mean $\gcd(|A|, |G|) = 1$? $\endgroup$ – Qiaochu Yuan Sep 29 at 18:41
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    $\begingroup$ What is a coprime action? $\endgroup$ – JCAA Sep 29 at 18:42
  • $\begingroup$ If the action is by automorphisms, the identity will always be a fixed point $\endgroup$ – ahulpke Sep 29 at 18:54
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    $\begingroup$ @ahulpke I think a fixed point free action of a group is usually defined to mean one in which only the identity is fixed i.e. $C_G(A) = 1$, and I am guessing that coprime means $(|A|,|G|)=1$. But the answer to the question is clearly no, coprime actions need not be fixed point free. $\endgroup$ – Derek Holt Sep 29 at 19:03
  • $\begingroup$ Yes, @QiaochuYuan, coprime action mean gdc(|A|,|G|) = 1 $\endgroup$ – Eliana C. Rodrigues Sep 29 at 19:03

As Derek Holt pointed out in the comments, the action may have fixed points, even if we don't count the identity. For example, take $G$ to be the quaternion group of order 8, and let $A = \mathbb Z/3\mathbb Z$ act by cycling $i, j$, and $k$. This fixes the central element $-1$.

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