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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?

Learnmore
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3 Answers3

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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

$\langle x,y:x^5=1,y^{101}=1,xy=y^{36}x\rangle.$

Michael Burr
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  • can you please explain how are you getting such maps? – Learnmore Mar 01 '15 at 17:26
  • It's called the semidirect product. Have you seen that product yet? – Michael Burr Mar 01 '15 at 17:27
  • No I have learnt direct product of groups – Learnmore Mar 01 '15 at 17:28
  • It is much more complicated to explain the semidirect product. I am trying to think of a way to answer this question, but the most natural explanations involve the semidirect product.

    One thing that you can see is that when the subgroup $\mathbb{Z}/5$ is not normal, then the group cannot be cyclic. However, one still must show that such a group exists.

     – Michael Burr Mar 01 '15 at 17:32
  • it would be very nice of you if you can give some alternative solution – Learnmore Mar 01 '15 at 17:41
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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

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Arpit Kansal
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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 $105$.

user1337
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