By Lagrange's theorem the possible order of the subgroups of the group will be 1 or 3 or 37 or 111

Now by Sylow's 1st theorem this group must have a subgroup of order 3 and as well as 37.

Then we can prove that every abelian group of order 111 is cyclic.

Now my question is:

Is there be any non-abelian group of order 111?

If there be a non abelian group of order 111 then the given statement is False otherwise the statement is True .



Jungnickel gave a nice result in On the Uniqueness of the Cyclic Group of Order $n$ that any group of order $n$ is cyclic iff $(n,\varphi(n))=1$, which addresses this problem quickly.

Since $(111,\varphi(111))=(111,72)=3$, there are noncyclic groups of order $111$. The first paragraph of the above paper shows how to construct a nonabelian group of order $n$, when $(n,\varphi(n))\neq 1$.

  • $\begingroup$ But Z_10 is cyclic although (10,φ(10))≠1....But according to the necessity of your statement (10,φ(10)) must be 1.....@Ben West $\endgroup$
    – indrajit
    Jun 18 '16 at 7:22
  • 1
    $\begingroup$ @IndrajitGhosh $\mathbb{Z}_n$ is cyclic for any $n$. If $(n,\varphi(n))=1$, $\mathbb{Z}_n$ is the only group of order $n$. If $(n,\varphi(n))\neq 1$, there is still a cyclic group of order $n$, but there are others as well. For instance, the dihedral group of order $10$ is nonabelian. $\endgroup$
    – Ben West
    Jun 18 '16 at 7:27

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