Normal subgroup equals kernel of homomorphism

Verbal statement
A subgroup of a group occurs as the fact about::kernel of a group homomorphism if and only if it is normal.

Symbolic statement
A subgroup $$N$$ of a group $$G$$ occurs as the kernel of a group homomorphism if and only if, for every $$g$$ in $$G$$, $$gNg^{-1} \subseteq N$$.

Kernel of a group homomorphism
A map $$\varphi: G \to H$$ is a homomorphism of groups if


 * $$\varphi(gh) = \varphi(g)\varphi(h)$$ for all $$g, h$$ in $$G$$
 * $$\varphi(e) = e$$
 * $$\varphi(g^{-1}) = (\varphi(g))^{-1}$$

The kernel of $$\varphi$$ is defined as the inverse image of the identity element under $$\phi$$.

Normal subgroup
For the purpose of this statement, we use the following definition of normality: a subgroup $$H$$ is normal in a group $$G$$ if $$H$$ contains each of its conjugate subgroups, that is, $$gNg^{-1} \subseteq N$$ for every $$g$$ in $$G$$.

Related facts
Closely related to this are the isomorphism theorems.


 * First isomorphism theorem
 * Second isomorphism theorem
 * Third isomorphism theorem

Kernel of homomorphism implies normal subgroup
Let $$\varphi: G \to H$$ be a homomorphism of groups. We first prove that the kernel (which we call $$N$$) of $$\phi$$ is a subgroup:


 * Identity element: Since $$\varphi(e) = e$$, $$e$$ is contained in $$N$$
 * Product: Suppose $$a, b$$ are in $$N$$. Then $$\varphi(a) = e$$ and $$\varphi(b) = e$$. Using the fact that $$\varphi(ab) = \varphi(a)\varphi(b)$$, we conclude that $$\varphi(ab) = e$$. Hence $$ab$$ is also in $$N$$.
 * Inverse: Suppose $$a$$ is in $$N$$. Then $$\varphi(a) = e$$. Using the fact that $$\varphi(a^{-1} = \varphi(a)^{-1}$$, we conclude that $$\varphi(a^{-1}) = e$$. Hence, $$a^{-1}$$ is also in $$N$$.

Now we need to prove that $$N$$ is normal. In other words, we must show that if $$g$$ is in $$G$$ and $$n$$ is in $$N$$, then $$gng^{-1}$$ is in $$N$$.

Since $$n$$ is in $$N$$, $$\phi(n) = e$$.

Consider $$\varphi(gng^{-1}) = \varphi(g)\varphi(n)\varphi(g^{-1}) = \varphi(g)\varphi(g^{-1}) = \varphi(gg^{-1}) = \varphi(e) = e$$. Hence, $$gng^{-1}$$ must belong to $$N$$.

Normal subgroup implies kernel of homomorphism
Let $$N$$ be a normal subgroup of a group $$G$$. Then, $$N$$ occurs as the kernel of a group homomorphism. This group homomorphism is the quotient map $$\varphi: G \to G/N$$, where $$G/N$$ is the set of cosets of $$N$$ in $$G$$.

The map is defined as follows:

$$\varphi(x) = xN$$

Notice that the map is a group homomorphism if we equip the coset space $$G/N$$ with the following structure:

$$(aN)(bN)=abN$$

This gives a well-defined group structure because, on account of $$N$$ being normal, the equivalence relation of being in the same coset of $$N$$ yields a congruence.

Explicitly:


 * 1) The map is well-defined, because if $$a' = an_1, b' = bn_2$$ for $$n_1,n_2 \in N$$, then $$a'b' = an_1bn_2 = ab(b^{-1}n_1bn_2) \in abN$$ (basically, we're using that $$bN = Nb$$).
 * 2) The image of the map can be thought of as a group because it satisfies associativity ($$((aN)(bN))(cN) = (aN)((bN)(cN))$$), has an identity element ($$N$$ itself), has inverses (the inverse of $$aN$$ is $$a^{-1}N$$)

Textbook references

 * , Page 82, Proposition 7