[proofplan] We prove both implications directly. First, if $N$ is the kernel of a homomorphism $\varphi: G \to K$, then conjugating any element of $N$ by any element of $G$ keeps its image equal to the identity of $K$, so $N$ is normal. Conversely, if $N$ is normal, we construct the [quotient group](/theorems/790) $G/N$ using left cosets and define the canonical projection $\pi: G \to G/N$ by $\pi(g) = gN$. Normality makes the coset multiplication well-defined, and the kernel of $\pi$ is exactly $N$. [/proofplan] [step:Show that the kernel of a homomorphism is closed under conjugation] Assume there exist a group $K$ with identity element $e_K$ and a [group homomorphism](/page/Group%20Homomorphism) $\varphi: G \to K$ such that $N = \ker \varphi$. To prove $N \trianglelefteq G$, we verify the conjugation criterion for normality: for every $g \in G$ and every $n \in N$, the element $gng^{-1}$ lies in $N$. Let $g \in G$ and $n \in N$. Since $N = \ker \varphi$, we have $\varphi(n) = e_K$. Since $\varphi$ is a group homomorphism, $\varphi(gng^{-1}) = \varphi(g)\varphi(n)\varphi(g^{-1})$. Using $\varphi(n) = e_K$ and $\varphi(g^{-1}) = \varphi(g)^{-1}$, this becomes $\varphi(gng^{-1}) = \varphi(g)e_K\varphi(g)^{-1} = e_K$. Thus $gng^{-1} \in \ker \varphi = N$. Since this holds for all $g \in G$ and $n \in N$, the subgroup $N$ is normal in $G$. [guided] Assume there exist a group $K$ with identity element $e_K$ and a group homomorphism $\varphi: G \to K$ such that $N = \ker \varphi$. We want to prove that $N$ is normal. For a subgroup, normality means that conjugation by elements of the ambient group preserves the subgroup: for every $g \in G$ and every $n \in N$, we must prove $gng^{-1} \in N$. Fix $g \in G$ and $n \in N$. Because $N$ is the kernel of $\varphi$, membership in $N$ is exactly the condition that the image under $\varphi$ is the identity element of $K$. Thus $\varphi(n) = e_K$. Now compute the image of the conjugate $gng^{-1}$. Since $\varphi$ is a group homomorphism, it preserves products and inverses, so $\varphi(gng^{-1}) = \varphi(g)\varphi(n)\varphi(g^{-1})$. The inverse-preserving property gives $\varphi(g^{-1}) = \varphi(g)^{-1}$, and the kernel condition gives $\varphi(n) = e_K$. Therefore $\varphi(gng^{-1}) = \varphi(g)e_K\varphi(g)^{-1} = e_K$. This says precisely that $gng^{-1} \in \ker \varphi$. Since $\ker \varphi = N$, we conclude that $gng^{-1} \in N$. Because $g$ and $n$ were arbitrary, $N$ is closed under conjugation by every element of $G$, hence $N \trianglelefteq G$. [/guided] [/step] [step:Construct the quotient group from a normal subgroup] Assume now that $N \trianglelefteq G$. Define $G/N$ to be the set of left cosets, $G/N := \{gN : g \in G\}$. Define a binary operation on $G/N$ by $(aN)(bN) := (ab)N$ for $a,b \in G$. We verify that this operation is well-defined. Suppose $aN = a'N$ and $bN = b'N$. Then there exist $n_1,n_2 \in N$ such that $a' = an_1$ and $b' = bn_2$. Hence $a'b' = an_1bn_2 = ab(b^{-1}n_1b)n_2$. Since $N \trianglelefteq G$, the element $b^{-1}n_1b$ lies in $N$. Since $N$ is a subgroup, $(b^{-1}n_1b)n_2 \in N$. Therefore $a'b' \in abN$, so $a'b'N = abN$. Thus the product of cosets does not depend on the chosen representatives. With this operation, $G/N$ is a group. The identity element is the coset $N = e_GN$, because for every $g \in G$ we have $(N)(gN) = (e_Gg)N = gN$ and $(gN)(N) = (ge_G)N = gN$. For each coset $gN \in G/N$, its inverse is $g^{-1}N$, since $(gN)(g^{-1}N) = (gg^{-1})N = e_GN = N$ and $(g^{-1}N)(gN) = (g^{-1}g)N = e_GN = N$. Associativity follows from associativity in $G$: for $a,b,c \in G$, $((aN)(bN))(cN) = ((ab)c)N = (a(bc))N = (aN)((bN)(cN))$. Therefore $K := G/N$ is a group. [guided] Assume $N \trianglelefteq G$. We want to turn the set of left cosets into a group, and the only possible multiplication compatible with representatives is $(aN)(bN) := (ab)N$ for $a,b \in G$. The point that must be checked is that this formula depends only on the cosets $aN$ and $bN$, not on the chosen representatives $a$ and $b$. Suppose $aN = a'N$ and $bN = b'N$. By equality of left cosets, there exist $n_1,n_2 \in N$ such that $a' = an_1$ and $b' = bn_2$. Then $a'b' = an_1bn_2 = ab(b^{-1}n_1b)n_2$. Normality is used exactly here: since $N \trianglelefteq G$ and $n_1 \in N$, the conjugate $b^{-1}n_1b$ lies in $N$. Because $N$ is a subgroup and $n_2 \in N$, the product $(b^{-1}n_1b)n_2$ also lies in $N$. Hence $a'b' \in abN$, so $a'b'N = abN$. Thus the product of cosets is well-defined. Now we verify the group laws. The identity element is the coset $N = e_GN$, because for every $g \in G$ we have $(N)(gN) = (e_Gg)N = gN$ and $(gN)(N) = (ge_G)N = gN$. The inverse of $gN$ is $g^{-1}N$, since $(gN)(g^{-1}N) = (gg^{-1})N = e_GN = N$ and $(g^{-1}N)(gN) = (g^{-1}g)N = e_GN = N$. Finally, associativity follows from associativity in $G$: for $a,b,c \in G$, $((aN)(bN))(cN) = ((ab)c)N = (a(bc))N = (aN)((bN)(cN))$. Therefore $G/N$ is a group under this coset multiplication. [/guided] [/step] [step:Define the quotient homomorphism and compute its kernel] Define the canonical quotient map $\pi: G \to G/N$ by $g \mapsto gN$. For $a,b \in G$, the definition of the operation on $G/N$ gives $\pi(ab) = (ab)N = (aN)(bN) = \pi(a)\pi(b)$. Thus $\pi$ is a group homomorphism. We now compute its kernel. By definition, $\ker \pi = \{g \in G : \pi(g) = N\}$. Since $\pi(g) = gN$, this becomes $\ker \pi = \{g \in G : gN = N\}$. For any $g \in G$, the equality $gN = N$ holds if and only if $g \in N$: if $gN = N$, then $g = ge_G \in gN = N$; conversely, if $g \in N$, then $gN = N$ because $N$ is a subgroup. Therefore $\ker \pi = N$. Taking $K := G/N$ and $\varphi := \pi$, we have found a group $K$ and a group homomorphism $\varphi: G \to K$ whose kernel is $N$. [guided] Now assume $N \trianglelefteq G$. The natural way to realize $N$ as a kernel is to collapse all elements of $N$ to the identity by passing to the quotient group. We define $K := G/N = \{gN : g \in G\}$. By the construction in the preceding step, normality of $N$ makes the multiplication $(aN)(bN) := (ab)N$ well-defined on left cosets, and this operation makes $G/N$ a group with identity coset $N = e_GN$. Define the map $\pi: G \to G/N$ by $g \mapsto gN$. We check that $\pi$ is a group homomorphism. Let $a,b \in G$. Then $\pi(ab) = (ab)N$. By the definition of multiplication in the quotient group, $(ab)N = (aN)(bN)$. Since $\pi(a) = aN$ and $\pi(b) = bN$, we get $\pi(ab) = \pi(a)\pi(b)$. Thus $\pi$ is a homomorphism. It remains to identify its kernel. The identity element of $G/N$ is the coset $N$, so $\ker \pi = \{g \in G : \pi(g) = N\}$. Substituting the definition $\pi(g) = gN$, this is $\ker \pi = \{g \in G : gN = N\}$. We now prove that $gN = N$ exactly when $g \in N$. If $gN = N$, then $g = ge_G$ belongs to $gN$, and since $gN = N$, this gives $g \in N$. Conversely, if $g \in N$, then left multiplication by $g$ permutes the subgroup $N$, so $gN = N$. More explicitly, $gN \subset N$ because $N$ is closed under multiplication, and $N \subset gN$ because for every $n \in N$ we have $n = g(g^{-1}n)$ with $g^{-1}n \in N$. Therefore $\ker \pi = N$. Taking $K = G/N$ and $\varphi = \pi$, we have constructed a group and a homomorphism whose kernel is exactly the given [normal subgroup](/page/Normal%20Subgroup) $N$. [/guided] [/step] [step:Combine the two implications] The first step proves that every subgroup which is the kernel of a group homomorphism is normal. The quotient construction proves that every normal subgroup $N \trianglelefteq G$ is the kernel of the canonical projection $\pi: G \to G/N$. Hence $N \trianglelefteq G$ if and only if there exist a group $K$ and a group homomorphism $\varphi: G \to K$ such that $N = \ker \varphi$. [/step]