[proofplan] The proof is a direct unpacking of the definition of a [simple group](/page/Simple%20Group). In the forward direction, simplicity forces every [normal subgroup](/page/Normal%20Subgroup) $N \trianglelefteq G$ to be either $\{e\}$ or $G$, and the corresponding quotient facts are verified directly from cosets. In the reverse direction, the stated quotient alternative already forces every normal subgroup to be either the identity subgroup or the whole group, which is exactly simplicity for the given nonidentity group. [/proofplan] [step:Verify the quotient alternatives when $G$ is simple] Assume that $G$ is simple. Let $N \trianglelefteq G$ be a normal subgroup. Since $G$ is nonidentity and simple, the only normal subgroups of $G$ are $\{e\}$ and $G$. Hence either $N=\{e\}$ or $N=G$. Suppose first that $N=\{e\}$. Define $\pi_{\{e\}}: G \to G/\{e\}$ by $\pi_{\{e\}}(g)=g\{e\}$. This map is the quotient homomorphism. It is surjective by definition of $G/\{e\}$ as the set of left cosets of $\{e\}$. If $g,h \in G$ satisfy $\pi_{\{e\}}(g)=\pi_{\{e\}}(h)$, then $g\{e\}=h\{e\}$, so $g=h$. Thus $\pi_{\{e\}}$ is injective. Since quotient multiplication is given by $(g\{e\})(h\{e\})=(gh)\{e\}$, the map $\pi_{\{e\}}$ is a bijective homomorphism, hence an isomorphism. Suppose next that $N=G$. Then every element $g \in G$ satisfies $gG=G$, so the [quotient group](/theorems/790) $G/G$ has exactly one element, namely the coset $G$. Define $\theta: G/G \to \{e\}$ by $\theta(G)=e$. This is a well-defined bijective homomorphism between one-element groups, so $G/G \cong \{e\}$. [guided] Assume that $G$ is simple, and let $N \trianglelefteq G$ be arbitrary. The definition of simplicity says that a nonidentity group has no normal subgroups except the identity subgroup and the whole group. Since the theorem already assumes $G$ is nonidentity, this applies directly: either $N=\{e\}$ or $N=G$. Now consider the first case, $N=\{e\}$. The quotient map is the function $\pi_{\{e\}}: G \to G/\{e\}$ defined by $\pi_{\{e\}}(g)=g\{e\}$ for each $g \in G$. It is a homomorphism because quotient multiplication satisfies \begin{align*} \pi_{\{e\}}(gh)=(gh)\{e\}=(g\{e\})(h\{e\})=\pi_{\{e\}}(g)\pi_{\{e\}}(h). \end{align*} It is surjective because every element of $G/\{e\}$ is a coset $g\{e\}$ for some $g \in G$. To prove injectivity, take $g,h \in G$ and assume $\pi_{\{e\}}(g)=\pi_{\{e\}}(h)$. Then $g\{e\}=h\{e\}$. Since $g\{e\}=\{g\}$ and $h\{e\}=\{h\}$, equality of these singleton sets gives $g=h$. Thus $\pi_{\{e\}}$ is a bijective homomorphism, so it is an isomorphism. Now consider the second case, $N=G$. For every $g \in G$, the coset $gG$ equals $G$, because left multiplication by $g$ permutes the elements of $G$. Hence the quotient group $G/G$ consists of exactly one element, the coset $G$. Define $\theta: G/G \to \{e\}$ by $\theta(G)=e$. The domain and codomain each have one element, so $\theta$ is bijective. It is a homomorphism because the product of the unique element of $G/G$ with itself is again the unique element, and the product of $e$ with itself in $\{e\}$ is $e$. Therefore $G/G \cong \{e\}$. [/guided] [/step] [step:Recover simplicity from the quotient criterion] Conversely, assume that for every normal subgroup $N \trianglelefteq G$, either $N=\{e\}$ and $\pi_N: G \to G/N$ is an isomorphism, or $N=G$ and $G/N \cong \{e\}$. Let $N \trianglelefteq G$ be any normal subgroup. By the assumed alternative, the first case gives $N=\{e\}$ and the second case gives $N=G$. Therefore every normal subgroup of $G$ is either $\{e\}$ or $G$. Since $G$ is nonidentity, this is exactly the definition of $G$ being simple. [guided] We now prove the reverse implication, so we assume the quotient alternative holds for every normal subgroup of $G$. To prove that $G$ is simple, we must show that an arbitrary normal subgroup of $G$ is forced to be either the identity subgroup or the whole group. Let $N \trianglelefteq G$ be any normal subgroup. The hypothesis applies to this particular $N$, and it gives exactly two possible alternatives. In the first alternative, $N=\{e\}$ and the quotient map $\pi_N: G \to G/N$ is an isomorphism. For the purpose of proving simplicity, the relevant conclusion from this alternative is the subgroup equality $N=\{e\}$. In the second alternative, $N=G$ and $G/N \cong \{e\}$. Again, for proving simplicity, the relevant conclusion is the subgroup equality $N=G$. Thus every normal subgroup $N \trianglelefteq G$ satisfies either $N=\{e\}$ or $N=G$. The theorem statement assumes that $G$ is nonidentity, so this normal-subgroup dichotomy is precisely the definition of $G$ being simple. Therefore $G$ is simple. [/guided] [/step]