Normal subgroup equals kernel of homomorphism

This article gives the statement, and possibly proof, of a basic fact in group theory.
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This article gives a proof/explanation of the equivalence of multiple definitions for the term normal subgroup
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Statement

Verbal statement

A subgroup of a group occurs as the 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 $G$, $gN{g}^{-1}\subseteq N$.

Definitions used

Kernel of a group homomorphism

A map $\phi :G\to H$ is a homomorphism of groups if

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

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

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, $gN{g}^{-1}\subseteq N$ for every $g$ in $G$.

Related facts

Closely related to this are the isomorphism theorems.

• First isomorphism theorem
• Second isomorphism theorem
• Third isomorphism theorem

Proof

Kernel of homomorphism implies normal subgroup

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

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

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

Consider $\phi (gn{g}^{-1})=\phi (g)\phi (n)\phi (g^{-1})=\phi (g)\phi (g^{-1})=\phi (g{g}^{-1})=\phi (e)=e$. Hence, $gn{g}^{-1}$ 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 $\phi :G\to G/N$, where $G/N$ is the set of cosets of $N$ in $G$.

The map is defined as follows:

$\phi (x)=xN$

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 $a^{\prime }=a{n}_1,b^{\prime }=b{n}_2$ for $n_1,n_2\in N$, then $a^{\prime }{b}^{\prime }=a{n}_{1}b{n}_2=ab(b^{-1}{n}_{1}b{n}_2)\in abN$ (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 $a^{-1}N$)

Further information: quotient map

References

Textbook references

• Abstract Algebra by David S. Dummit and Richard M. Foote, 10-digit ISBN 0471433349, 13-digit ISBN 978-0471433347, ^{More info}, Page 82, Proposition 7