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Let $p$ be prime. If a group has more than $p-1$ elements of order $p$, why can't the group be cyclic?

Can I assume that the group must be finite group since there is at least one element with finite order?

And I am wondering now if all cyclic groups are finite, or all finite groups are cyclic?

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marked as duplicate by Dietrich Burde abstract-algebra Mar 14 '18 at 14:13

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  • $\begingroup$ Answer to the last question : No, not all finite groups are cyclic. And $(\mathbb Z,+)$ is cyclic, but not finite. $\endgroup$ – Peter Mar 14 '18 at 14:10
  • $\begingroup$ Also, not all cyclic groups are finite: $\mathbb{Z} = \langle 1 \rangle$ $\endgroup$ – JackR Mar 14 '18 at 14:10
  • $\begingroup$ All finite groups are not cylic . For example $\Bbb Z_4, D_8, ...$. All cylic groups are not finite.For example $\Bbb Z=\langle 1\rangle$. $\endgroup$ – 1ENİGMA1 Mar 14 '18 at 14:12
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    $\begingroup$ $\mathbb Z_4$ however is cyclic. $\endgroup$ – Peter Mar 14 '18 at 14:13
  • $\begingroup$ Hint. Look at the subgroup generated by an element of order $p$. (The integers under addition are an infinite cyclic group. The six element group of symmetries of a triangle isn't cyclic.) $\endgroup$ – Ethan Bolker Mar 14 '18 at 14:13
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An infinite cyclic group has no elements of finite order, except for $e$, which has order $1$. Thus, your group cannot be infinite cyclic.

A finite cyclic group of order $n$ has exactly one subgroup of order $d$ for each $d$ dividing $n$.

A subgroup of order $p$ in any group has exactly $p-1$ elements of order $p$.

Therefore, if a group has more than $p-1$ elements of order $p$, then it has more than one subgroup of order $p$ and so cannot be cyclic.

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  • $\begingroup$ "A subgroup of order p in any group has exactly $p−1$ elements of order $p$ ." Is this because an element of the order must divide the order of its group? And since the group's order is prime, we can't find the element with the order other than 1 or $p$. The element of order 1 must be identity, so we are left with $p-1$ many elements who must have order $p$? $\endgroup$ – user3000482 Mar 14 '18 at 16:45
  • $\begingroup$ @user3000482, exactly $\endgroup$ – lhf Mar 14 '18 at 19:39

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