[proofplan] We prove both implications directly from the definition of a [simple group](/page/Simple%20Group). In the forward direction, the kernel of any homomorphism out of $G$ is a [normal subgroup](/page/Normal%20Subgroup) of $G$, so simplicity forces it to be either the trivial subgroup or all of $G$. In the reverse direction, every normal subgroup $N \trianglelefteq G$ appears as the kernel of the canonical quotient homomorphism $\pi: G \to G/N$, so the assumed kernel condition forces $N$ to be either $\{e_G\}$ or $G$. [/proofplan]

[step:Show that kernels of homomorphisms out of a simple group have only the two possible sizes]Assume $G$ is simple. Let $H$ be a group with identity element $e_H$, and let \begin{align*} \varphi: G \to H \end{align*} be a [group homomorphism](/page/Group%20Homomorphism). Define \begin{align*} \ker \varphi := \{g \in G : \varphi(g)=e_H\}. \end{align*} We verify that $\ker \varphi \trianglelefteq G$. If $a,b \in \ker \varphi$, then \begin{align*} \varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1}=e_H e_H^{-1}=e_H, \end{align*} so $ab^{-1} \in \ker \varphi$; hence $\ker \varphi \le G$. If $g \in G$ and $k \in \ker \varphi$, then \begin{align*} \varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}=\varphi(g)e_H\varphi(g)^{-1}=e_H, \end{align*} so $gkg^{-1} \in \ker \varphi$. Therefore $\ker \varphi \trianglelefteq G$. Since $G$ is simple and $G \ne \{e_G\}$, the only normal subgroups of $G$ are $\{e_G\}$ and $G$. Hence $\ker \varphi=\{e_G\}$ or $\ker \varphi=G$.[/step]

[guided]Assume $G$ is simple. To prove the kernel condition, we must start with an arbitrary target group and an arbitrary homomorphism out of $G$. Let $H$ be a group with identity element $e_H$, and let \begin{align*} \varphi: G \to H \end{align*} be a group homomorphism. The kernel is the subset \begin{align*} \ker \varphi := \{g \in G : \varphi(g)=e_H\}. \end{align*} The definition of simplicity applies to normal subgroups, so the key point is to verify that $\ker \varphi$ is a normal subgroup of $G$. First we check that it is a subgroup. If $a,b \in \ker \varphi$, then $\varphi(a)=e_H$ and $\varphi(b)=e_H$. Since $\varphi$ is a homomorphism, \begin{align*} \varphi(ab^{-1}) = \varphi(a)\varphi(b)^{-1}=e_H e_H^{-1}=e_H. \end{align*} Thus $ab^{-1}\in \ker \varphi$, so the [subgroup criterion](/theorems/932) gives $\ker \varphi \le G$. Now we check normality. Let $g \in G$ and $k \in \ker \varphi$. Since $\varphi(k)=e_H$ and $\varphi$ preserves multiplication and inverses, \begin{align*} \varphi(gkg^{-1})=\varphi(g)\varphi(k)\varphi(g)^{-1}=\varphi(g)e_H\varphi(g)^{-1}=e_H. \end{align*} Therefore $gkg^{-1}\in \ker \varphi$ for every $g \in G$ and every $k \in \ker \varphi$, which is precisely the conjugation-invariance condition for normality. Hence $\ker \varphi \trianglelefteq G$. Because $G$ is simple and nonidentity, its only normal subgroups are $\{e_G\}$ and $G$. Applying this to the normal subgroup $\ker \varphi$, we obtain \begin{align*} \ker \varphi=\{e_G\} \end{align*} or \begin{align*} \ker \varphi=G. \end{align*} This proves the forward implication.[/guided]

[step:Recover every normal subgroup as the kernel of its quotient map]Assume that for every group $H$ and every group homomorphism $\varphi: G \to H$, one has $\ker \varphi=\{e_G\}$ or $\ker \varphi=G$. Let $N \trianglelefteq G$ be a normal subgroup. Since $N$ is normal, the quotient set $G/N$ carries the [quotient group](/theorems/790) structure whose elements are left cosets $gN$. Define the canonical quotient homomorphism \begin{align*} \pi: G \to G/N \end{align*} by \begin{align*} \pi(g)=gN. \end{align*} For $g,h \in G$, normality of $N$ makes coset multiplication well-defined, and \begin{align*} \pi(gh)=ghN=(gN)(hN)=\pi(g)\pi(h), \end{align*} so $\pi$ is a group homomorphism. The identity element of $G/N$ is $N=e_GN$. Therefore \begin{align*} \ker \pi=\{g \in G : gN=N\}. \end{align*} We prove this set is exactly $N$. If $g \in N$, then $gN=N$. Conversely, if $gN=N$, then $g=ge_G \in gN=N$, so $g \in N$. Hence \begin{align*} \ker \pi=N. \end{align*}[/step]

[guided]Assume the kernel condition: for every group $H$ and every homomorphism $\varphi: G \to H$, the kernel is either $\{e_G\}$ or $G$. To prove that $G$ is simple, we must show that every normal subgroup of $G$ is one of these two subgroups. Let $N \trianglelefteq G$ be a normal subgroup. The reason normality is exactly the right hypothesis is that it permits the construction of the quotient group $G/N$. Its elements are cosets $gN$ with $g \in G$, and multiplication is defined by \begin{align*} (gN)(hN)=ghN. \end{align*} This multiplication is well-defined because $N$ is normal. Now define the canonical quotient map \begin{align*} \pi: G \to G/N \end{align*} by \begin{align*} \pi(g)=gN. \end{align*} We verify that this is a group homomorphism. For $g,h \in G$, the definition of multiplication in the quotient group gives \begin{align*} \pi(gh)=ghN=(gN)(hN)=\pi(g)\pi(h). \end{align*} Thus $\pi$ is one of the homomorphisms to which the assumed kernel condition applies. It remains to identify its kernel. The identity element of $G/N$ is the coset $N=e_GN$, so \begin{align*} \ker \pi=\{g \in G : \pi(g)=N\}=\{g \in G : gN=N\}. \end{align*} We show that this set is precisely $N$. If $g \in N$, then multiplying $N$ on the left by $g$ does not change the coset, so $gN=N$. Conversely, if $gN=N$, then $g=ge_G$ belongs to $gN$, and since $gN=N$, this gives $g \in N$. Therefore \begin{align*} \ker \pi=N. \end{align*} So every normal subgroup $N$ of $G$ is realised as the kernel of a group homomorphism out of $G$.[/guided]

[step:Apply the kernel condition to the quotient map and conclude simplicity] By the assumed kernel condition applied to the homomorphism $\pi: G \to G/N$, we have $\ker \pi=\{e_G\}$ or $\ker \pi=G$. Since $\ker \pi=N$, it follows that $N=\{e_G\}$ or $N=G$. Thus every normal subgroup of $G$ is either $\{e_G\}$ or $G$. Because $G \ne \{e_G\}$ by hypothesis, this is exactly the definition that $G$ is simple. This proves the reverse implication and completes the proof. [/step]