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The Corollary: If $G$ is a prime of order $p$, then $G$ is cyclic.

The Proof: Let $ x \in G$, $x \neq 1$. Thus $|\langle x\rangle| > 1$ and $|\langle x \rangle|$ divides $|G|$. Since $|G|$ is prime we must have $|\langle x \rangle| = |\langle G \rangle|$, hence $G = \langle x \rangle$ is cyclic.

My problem with the proof, is how do you know that any random $x$ in $G$, will generate a subgroup. For example if your group was $\langle Z, +\rangle$ then $\langle1\rangle$ doesn't produce a subgroup of $Z$.

Ali Caglayan
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user123429
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    $\langle 1\rangle$ does produce a subgroup of $\mathbb Z$ - the whole group is still a subgroup. Note that this corollary can only apply to finite groups. – Mathmo123 Oct 11 '14 at 22:59
  • ahh your right, I forgot you can include the inverse, I was thinking it was 1, 2, 3, 4,.... So if you are to take any element in a group, will generating that element always give you a subgroup? – user123429 Oct 11 '14 at 23:01
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    Yes. Or indeed if you take any set of elements, you can look at the group spanned by those elements. $\langle x \rangle$ can be thought of as the smallest subgroup of $G$ containing $x$ – Mathmo123 Oct 11 '14 at 23:03

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Let $G$ be a group of prime order and let $x \in G$, $x \neq e$. As $x \neq e$, then $x^2 \neq x$, and so the group $\langle x \rangle = \{x^k \ | \ k \in \mathbb{N}\}$ has at least two elements.

By Lagrange's Theorem: $ o(\langle x \rangle)$ divides $o(G) $, and then, because $o(G)$ is prime, $o(\langle x \rangle) = o(G)$ and then $\langle x \rangle = G$ and $G$ is cyclic.

Jonas Gomes
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