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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?

I understand how to prove this if the group is finite because the contrapositive of this statement is true due to the Euler-$\phi$ function $\phi (p) = p - 1$, which is the number of elements of order $p$.

But how would I prove this for infinite groups?

Oliver G
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    Are you aware that there is only one infinite cyclic group? – Milo Brandt Jul 16 '16 at 14:39
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    I am unaware of that. I can see that $\langle 1 \rangle$ and $\langle -1 \rangle$ generate $\Bbb Z$, and $\langle k \rangle$ and $\langle -k \rangle$ generate $k \Bbb Z$, but I haven't proven that there are no other infinite cyclic groups. – Oliver G Jul 16 '16 at 14:49
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    Given any infinite cyclic group $\langle g \rangle$, you get an isomorphism by mapping $g$ to $1\in \mathbb{Z}$. – J126 Jul 16 '16 at 14:50

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The only element of finite order in an infinite cyclic group is the neutral element, which has order $1$. (If you do not know this, prove it.) Thus, for an infinite cyclic group there are never more than $p-1$ elements of order $p$ for $p$ a prime; indeed there never exists an element of order $p$.

quid
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