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I've been asked to prove this.

In class we proved this when $n=15$, but our approached seemed unnecessarily complicated to me. We invoked Sylow's theorems, normalizers, etc. I've looked online and found other examples of this approach.

I wonder if it is actually unnecessary, or if there is something wrong with the following proof:

If $|G|=35=5\cdot7$ , then by Cauchy's theorem, there exist $x,y \in G$ such that $o(x)=5$, $o(y)=7$. The order of the product $xy$ is then $\text{lcm}(5,7)=35$. Since we've found an element of $G$ of order 35, we conclude that $G$ is cyclic.

Thanks.

  • 0
    I can't see what's wrong with it, but I have a hard time believing that so many people would overlook this proof it it were correct.2011-10-26
  • 25
    If $x$ and $y$ commute then the order of $xy$ is the lcm, but you can't assume this.2011-10-26
  • 1
    In fact, even if $x$ and $y$ have finite order, it need not be that $xy$ has finite order.2011-10-26
  • 0
    (following DJC's comment) e.g. We can express the Fibonacci matrix (with infinite order) as the product of two matrices with finite order (the orders are 2 and 3, if memory serves).2011-10-26
  • 6
    So try applying this to a group of order 21. What goes wrong (since there is a non-abelian group of that order, which is therefore non-cyclic - as every cyclic group is abelian)? If you can't distinguish the cases, you don't have a proof.2011-10-26
  • 0
    The order of the product being $35$ only works if you know $G$ is commutative.2011-10-26

7 Answers 7