[proofplan] The key point is that in an [abelian group](/page/Abelian%20Group) every subgroup is normal, so simplicity turns any nonidentity subgroup into the whole group. Starting from a nonidentity element, this forces a simple abelian group to be cyclic. We then show directly that a cyclic [simple group](/page/Simple%20Group) cannot have infinite or composite finite order, so its order is prime. Conversely, a [cyclic group](/page/Cyclic%20Group) of prime order has no nontrivial proper subgroup, hence no nontrivial proper [normal subgroup](/page/Normal%20Subgroup). [/proofplan] [step:Force a simple abelian group to be cyclic] Assume first that $G$ is simple and abelian. Since a simple group is nonidentity, choose an element $g \in G$ with $g \ne e$. Define the cyclic subgroup generated by $g$ by \begin{align*} \langle g\rangle := \{g^m : m \in \mathbb{Z}\}. \end{align*} Here $\mathbb{Z}$ denotes the additive group of integers, used as the exponent set. This is a subgroup of $G$, and it is nonidentity because $g \in \langle g\rangle$ and $g \ne e$. Since $G$ is abelian, every subgroup of $G$ is normal: if $H \le G$, $x \in G$, and $h \in H$, then $x h x^{-1} = h x x^{-1} = h \in H$, so $xHx^{-1} \subset H$, and applying the same argument to $x^{-1}$ gives equality. Hence $\langle g\rangle \trianglelefteq G$. By simplicity of $G$, the nonidentity normal subgroup $\langle g\rangle$ must equal $G$. Therefore $G$ is cyclic, generated by $g$. [guided] We begin by using the only special feature of abelian groups needed in the proof: their subgroups are automatically normal. Since $G$ is simple, by the standard convention $G$ is not the identity group. Hence there exists an element $g \in G$ such that $g \ne e$, where $e$ is the identity element of $G$. Define the subgroup generated by $g$ as \begin{align*} \langle g\rangle := \{g^m : m \in \mathbb{Z}\}. \end{align*} Here $\mathbb{Z}$ denotes the additive group of integers, used as the exponent set. This set is a subgroup of $G$: it contains $e = g^0$, it is closed under products because $g^m g^n = g^{m+n}$ for $m,n \in \mathbb{Z}$, and it is closed under inverses because $(g^m)^{-1} = g^{-m}$. It is nonidentity because $g = g^1$ belongs to $\langle g\rangle$ and $g \ne e$. Now we verify normality. Let $H \le G$ be any subgroup, let $x \in G$, and let $h \in H$. Since $G$ is abelian, $xh = hx$, so \begin{align*} x h x^{-1} = h x x^{-1} = h. \end{align*} Thus $xHx^{-1} \subset H$. Replacing $x$ by $x^{-1}$ gives the reverse inclusion, so $xHx^{-1} = H$ for every $x \in G$. Therefore every subgroup of $G$ is normal. Applying this to $H = \langle g\rangle$, we have $\langle g\rangle \trianglelefteq G$. Simplicity says that the only normal subgroups of $G$ are $\{e\}$ and $G$. Since $\langle g\rangle$ is not $\{e\}$, it follows that $\langle g\rangle = G$. Thus every element of $G$ is a power of $g$, so $G$ is cyclic. [/guided] [/step] [step:Exclude infinite and composite order for the cyclic group] Since $G$ is cyclic, choose a generator $a \in G$ with $G = \langle a\rangle$. First suppose $a$ has infinite order. Define \begin{align*} K := \langle a^2\rangle = \{a^{2m} : m \in \mathbb{Z}\}. \end{align*} Then $K \le G$ and $K$ is nonidentity because $a^2 \ne e$. Also $K \ne G$, since $a \notin K$: if $a = a^{2m}$ for some $m \in \mathbb{Z}$, then $a^{2m-1} = e$, contradicting that $a$ has infinite order. Because $G$ is abelian, $K \trianglelefteq G$, contradicting simplicity. Hence $a$ has finite order. Let $n := \operatorname{ord}(a) = |G|$. If $n$ were composite, choose a [prime number](/page/Prime%20Number) $q$ with $q \mid n$ and $q < n$. Define \begin{align*} L := \langle a^q\rangle = \{a^{qm} : m \in \mathbb{Z}\}. \end{align*} Then $L \le G$. The element $a^q$ is not the identity, because $0 < q < n$ and $n$ is the least positive integer with $a^n=e$. Thus $L$ is nonidentity. Also $L \ne G$, because if $a \in L$, then $a = a^{qm}$ for some $m \in \mathbb{Z}$, so $a^{qm-1}=e$, and therefore $n \mid qm-1$. Since $q \mid n$, this would imply $q \mid qm-1$, impossible. Again $L$ is normal because $G$ is abelian, contradicting simplicity. Therefore $n$ is prime. [/step] [step:Identify the group as cyclic of prime order] From the previous step, $G = \langle a\rangle$ and $|G| = n$ for a prime number $n$. Put $p := n$. By definition, every cyclic group of order $p$ is isomorphic to $C_p$. Hence \begin{align*} G \cong C_p. \end{align*} This proves the forward implication. [/step] [step:Show that a cyclic group of prime order is simple] Conversely, assume $G \cong C_p$ for a prime number $p$. Since simplicity is invariant under [group isomorphism](/page/Group%20Isomorphism), it suffices to prove that $C_p$ is simple. Let $c \in C_p$ be a generator, so \begin{align*} C_p = \langle c\rangle \end{align*} and $\operatorname{ord}(c)=p$. Let $H \le C_p$ be a subgroup. If $H=\{e\}$, there is nothing to prove. Otherwise choose $x \in H$ with $x \ne e$. Since $C_p = \langle c\rangle$, there exists an integer $k$ with $x=c^k$. The condition $x \ne e$ means $p \nmid k$. Since $p$ is prime and $p \nmid k$, the [greatest common divisor](/page/Greatest%20Common%20Divisor) satisfies $\gcd(k,p)=1$. Bezout's identity gives integers $m$ and $r$ such that $mk+rp=1$, hence $mk \equiv 1 \pmod p$. Therefore \begin{align*} c = c^{mk} = (c^k)^m = x^m. \end{align*} Because $x \in H$ and $H$ is a subgroup, $x^m \in H$, so $c \in H$. Therefore $H=C_p$. Thus the only subgroups of $C_p$ are $\{e\}$ and $C_p$. Since $C_p$ is abelian, every subgroup is normal, so the only normal subgroups are $\{e\}$ and $C_p$. Also $C_p$ is nonidentity because $p$ is prime and hence $p \ge 2$. Therefore $C_p$ is simple. Consequently every group isomorphic to $C_p$ is simple, completing the proof. [/step]