Suppose we want to show that group of order $72$ is solvable. Why

Definition: The group $G$ is called solvable if there exists abelian normal tower, i.e. $$\{e\}=G_0\vartriangleleft G_1\vartriangleleft\dots \vartriangleleft G_n=G,$$ such that $G_{i+1}/G_{i}$ is abelian group.

I know the method of showing that any group of order $72$ is NOT simple.

In order to show that any group of order $72$ is solvable why it's enough to show that it NOT simple?

Would be very grateful for detailed explanation. In some sense I am not able to understand the relation between simplicity and solvability.

  • 2
    $\begingroup$ It's well-known that non-Abelian groups of order $<60$ are never simple. $\endgroup$ – Lord Shark the Unknown Nov 7 '18 at 4:23
  • $\begingroup$ @LordSharktheUnknown, is there any alternative explanation? $\endgroup$ – ZFR Nov 7 '18 at 4:26
  • $\begingroup$ Burnside p^aq^b? $\endgroup$ – user10354138 Nov 7 '18 at 4:28
  • $\begingroup$ As a problem, showing that a group of order 72 is soluble only makes sense if you have already proved that groups of smaller order--dividing 72 properly--are also soluble. $\endgroup$ – the_fox Nov 7 '18 at 4:32
  • $\begingroup$ Consider Lord Shark's hint together with the useful fact that if $H\unlhd G$, then $G$ is solvable if and only if $H$ and $G/H$ are both solvable. $\endgroup$ – Fimpellizieri Nov 7 '18 at 5:55

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