On the Frobenius group

Question

I am studying some properties of Frobenius group $G$ of order $20$, which mean that a presentation of the group $G$ is

$$G = \langle c, f \mid c^5 = f^4 = 1, \,cf = fc^2\rangle.$$

My question is that true $f^2c = cf^2$? I use the following simple proof:

$$\begin{align} cf = fc^2 &\Rightarrow fc = c^2f\\ &\Rightarrow ffc = fc^2f\\ & \Rightarrow f^2c = cff\\ & \Rightarrow f^2c = cf^2. \end{align}$$

I think Its seems trivial but I need confirmation because the given reasoning help to get more relations betwen $c$ and $f$.

Answer 1

$cf=fc^2⇒fc=c^2f$. This step is false. You can take inverses of the first words to get $f^{-1}c^{-1}=c^{-2}f^{-1}$, but you cannot just reverse words. In fact, no (non-trivial) power of $c$ commutes with any (non-trivial) power of $f$.

Answer 2

The relation $f^2 c = c f^2$ would imply that $$f^{-2} c f^2 = f^{-1} (f^{-1} c f) f = f^{-1} c^2 f = (f^{-1} c f)^2 = c^4$$ is equal to $c$. But $c^3 = 1$ and $c^5 = 1$ force $c = 1$, which is not the case in $G$. (This is immediate if you know that your group has order $20$, since the condition $c = 1$ would then force $G = \langle{f:\, f^5 = 1\rangle} = \mathbb{Z}_5$. Without that, we could use the presentation of $G$ to define a suitable homomorphism onto some group of order larger than $5$.)

The problem in your proof is that $xy = zw$ does not imply $yx = wz$; in fact, $xy = w$ does not imply $yx = w$. The group $G$ is not abelian, and in particular is not just $\mathbb{Z}_5 \oplus \mathbb{Z}_4$.