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  "body": "\n## Proof\n\nWe prove the two directions, then the last claim.\n\n1. Let $G$ be [[abelian|071DDFVG4J]] and [[simple|0C093D1T4M]]. By definition $G \\neq \\{e\\}$, so we can pick $g \\in G$ with $g \\neq e$.\n2. Each subgroup of $G$ is normal, because [[subgroups of abelian groups are normal|0JPDTEDCFD]]. So the only subgroups of $G$ are $\\{e\\}$ and $G$.\n3. The [[cyclic subgroup|0YT586TKXF]] $\\langle g \\rangle$ contains $g \\neq e$, so by step 2 it equals $G$.\n4. Suppose that $g$ has infinite order. Then the powers $g^k$, $k \\in \\Z$, are distinct, by [[this result|031M8HX7DA]]. So $g^2 \\neq e$.\n5. In that case $g \\notin \\langle g^2 \\rangle$. Otherwise $g = g^{2k}$ for some $k \\in \\Z$, and the powers $g^1$ and $g^{2k}$ with $1 \\neq 2k$ would be equal.\n6. So $\\langle g^2 \\rangle$ is a subgroup other than $\\{e\\}$ and $G$, which contradicts step 2. Hence $g$ has finite order $n$.\n7. By [[this result|031M8HX7DA]] and step 3, $|G| = |\\langle g \\rangle| = n$, and $n \u003e 1$ because $g \\neq e$.\n8. Suppose that $n$ is not prime. Then $n = ab$ with integers $1 \u003c a \u003c n$ and $1 \u003c b \u003c n$.\n9. Let $x = g^a$. Then $x^b = g^n = e$. For $0 \u003c j \u003c b$ we have $0 \u003c aj \u003c n$, so $x^j = g^{aj} \\neq e$. So $x$ has order $b$.\n10. By [[this result|031M8HX7DA]], $|\\langle x \\rangle| = b$ with $1 \u003c b \u003c n$. So $\\langle x \\rangle$ is neither $\\{e\\}$ nor $G$, which contradicts step 2.\n11. Hence $G$ is finite and its order $n$ is a prime number. This proves the first direction.\n12. Conversely, let $G$ be abelian of prime order $p$. Then $G \\neq \\{e\\}$.\n13. Let $N$ be a normal subgroup of $G$. By [[Lagrange's theorem|3C53TD]], $|N|$ divides $p$, so $|N| = 1$ or $|N| = p$. That is, $N = \\{e\\}$ or $N = G$. So $G$ is simple.\n14. Finally, let $G$ be abelian and simple, so $|G| = p$ is prime by step 11. By [[groups of prime order are cyclic|05MM4EK4SA]], $G$ is [[cyclic|0B85Q41RP6]].\n15. The group $\\Z/p\\Z$ is cyclic of order $p$, generated by the class of $1$. By [[this result|0736AXH2ME]], $G \\cong \\Z/p\\Z$.\n\n## Idea\n\nIn an abelian group, every subgroup is normal, so simplicity means there are no subgroups at all besides the trivial ones. Any element $g \\neq e$ then generates the whole group. A proper power of $g$ would generate a smaller nontrivial subgroup unless the order of $g$ is prime.\n",
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