# Simplicity of group of order $72$ follows to solvability?

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.

• It's well-known that non-Abelian groups of order $<60$ are never simple. – Lord Shark the Unknown Nov 7 '18 at 4:23
• @LordSharktheUnknown, is there any alternative explanation? – ZFR Nov 7 '18 at 4:26
• Burnside p^aq^b? – user10354138 Nov 7 '18 at 4:28
• 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. – the_fox Nov 7 '18 at 4:32
• 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. – Fimpellizieri Nov 7 '18 at 5:55