Denote by $P_5$ the cyclic Sylow 5-subgroup (since the number Sylow 5-subgroup is one). Let $X$ be the generating set of $P_5$, i.e. $X=\{g\in G\mid P_5=\langle g \rangle\}$. We need to show Sylow 3-subgroups act on $X$ by conjugation, and give possible numbers of the size of orbits of this action.

I know that if $n_3=1$ ($n_3$ is the number of Sylow 3-subgroups), then $P_3$ fixes every points of $X$, and $G=P_3P_5$ is cyclic. What will happen if $n_3=25$, and what is the relationship between this group action and $G$ being cyclic?

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    $\begingroup$ Proper notation is $P_5 = \langle g\rangle,$ not $P_5 = <g>. \qquad$ $\endgroup$ – Michael Hardy Jan 7 '18 at 21:33

Let $\;P_3=\langle x\rangle\;$ a Sylow $\;3\,-$ subgroup , and define and action

$$\;P_3\times P_5=\langle g\rangle\to P_5\;,\;\;g^x:=x^{-1}gx$$

It must be clear this indeed is an action of $\;P_3\;$ by automorphisms on $\;P_5\;$ .

But $\;\left|\text{Aut}(P_5)\right|=\phi(125)=5^2(5-1)=100\;$ , and this means that the homomorphism $\;P_3\to\text{Aut}(P_5)\;$ defined by the above action must be the trivial one, which means $\;P_3\;$ centralizes $\;P_5\;$ (which in this case simply means $\;G\;$ is abelian...), and thus $\;G=P_3P_5=P_3\times P_5\;$


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