Normal subgroup equals kernel of homomorphism

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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 $G$, $gNg^{-1} \subseteq N$.

Definitions used

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.

Proof

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 $a^{-1}N$)

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