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I'm reading through Contemporary Abstract Algebra by Joseph A. Galian and I stumbled upon the following problem:

Suppose $G$ is an Abelian group of order $35$ and every element of $G$ satisfies $x^{35} = e$. Prove that $G$ is cyclic.

My approach was the following:
I divided the problem into these three cases:

  1. $G$ has elements of order $5$ and elements of order $7$
  2. Every non-trivial element of $G$ has order $5$
  3. Every non-trivial element of $G$ has order $7$

My idea was that if I manage to prove in the first case we will have an element of order $35$ and that the remaining two cases are impossible, I would have proved that $G$ must have at least an element of order $35$ which will make $G$ cyclic.

I dealt the first case as follows: Let $a\in G$ and $b\in G$ such that $\lvert a\rvert = 5$ and $\lvert b\rvert = 7$. In that case, $(ab)^5 = b^5 \not = e$ and $(ab)^7 = a^2 \not = e$, which means that $\lvert ab\rvert = 35$ hence $G =\lt ab \gt$.

However, I got stuck in the remaining two cases. Thank you in advance.

Shaun
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Selim Bamri
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    If you can use Sylow's Theorems (or Cauchy's Theorem), you can assure there is at least an element of order $5$ and another of order $7$, that is, we're always in the first case. – Julio Puerta Mar 02 '24 at 11:20
  • Thank you for your response, but at this stage I haven't gone through those theorems yet. Isn't there a more elementary way to prove that? – Selim Bamri Mar 02 '24 at 11:23
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    If every non-identity element of $G$ has order $5$ then there would be $(|G|-1)/4$ subgroups of order $5$, but $4$ does not divide $34$ so that's impossible, Similarly they cannot all have order $7$ because $6$ does not divide $34$. – Derek Holt Mar 02 '24 at 11:29
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    You can try to prove that $G\cong C_5\times C_7$, which is cyclic. – Deif Mar 02 '24 at 11:44
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    $$\text{Please don't type like this!}$$ – Shaun Mar 02 '24 at 14:12

1 Answers1

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The number of elements of order $p$, in any finite group, is divisible by $p-1.$

That's enough to establish elements of orders $5,7.$

calc ll
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