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Let $G$ be a finite group (possibly not commutative) and $d$ be an arbitrary number which divides $|G|$. If $G$ is square-free, i.e. $\forall n\colon n^2$ doesn't divide $ |G|$, can we say that $G$ has a subgroup of order $d$?

My first attempt was to use the Sylow's theorem. I denoted $d= p_1 p_2 \cdots p_n$ and took the Sylow $p_i$-subgroup $H_i$. However, I couldn't say that $H_1 H_2\cdots H_n$ is even a subgroup, since they are in general not normal subgroups.

I also tried to say that $G$ is nilpotent (so $G$ is the direct sum of $H_i$). I guess that's not true...

Bernard
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nessy
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    Yes, this is true, but it is not quite trivial to see. The crucial connection is that groups of squarefree order are supersolvable, and supersolvable groups satisfy the converse of Lagrange. – Tobias Kildetoft May 18 '20 at 18:39
  • Thank you for your help! I'll check it. – nessy May 18 '20 at 19:02

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