Let $G$ be a group of order 505.Is it cyclic? Let $G$ be a group of order 505.Is it cyclic?
My try:By Sylow's theorem $G$ has a subgroup of order 101 and a subgroup of order 5.If both have been unique we could have concluded that $G$ is cyclic.
But here no. of Sylow 5 subgroup =$1+5k$ divides 101 which holds for $k=0 $ and $k=20$
How to proceed here?
 A: From the Sylow theorems, you know that the group of order 101 is unique.  Since that subgroup is unique, it is normal.  Therefore, $G$ has a normal subgroup of order 101 and a subgroup of order $5$ (which might not be normal).  If you continue with the Sylow theorems, you will find that the groups of this type are called semidirect products determined by maps from $\mathbb{Z}/5$ into Aut$(\mathbb{Z}/101)\simeq\mathbb{Z}/100$.  There is a map from $\mathbb{Z}/5$ into $\mathbb{Z}/100$, and this gives a noncyclic group of order 505.    In fact, the group can be presented as
$\langle x,y:x^5=1,y^{101}=1,xy=y^{36}x\rangle.$
A: Suppose $G$ is a group of order $pq (p<q)$ where $p$ and $q$ are prime and $p$ divides $q-1$ then there always exists a unique nonabelian group of order $pq$.Can you conclude now?
Proof of above result can be found here :Non Abelian group of order pq
A: To expand on Michael Burr's answer, a number $n \in \mathbb{Z}^+$ is called a cyclicity-forcing  number if all groups of order $n$ are cyclic. There is a very nice characterization of these numbers, which can be found here. Since $n=105$ doesn't follow the criteria, there exists a non-cyclic group of order $105$.
