Normal subgroup

A normal subgroup ${\rm H}$ of a group ${\rm G}$ is a subgroup of ${\rm G}$ for which the relation " $xy^{-1} \in {\rm H}$ " of $x$ and $y$ is compatible with the law of composition on ${\rm G}$ , which in this article is written multiplicatively. The quotient group of ${\rm G}$ under this relation is often denoted ${\rm G/H}$ (said, " ${\rm G}$ mod ${\rm H}$ "). (Hence the notation $\mathbb{Z}/n\mathbb{Z}$ for the integers mod $n$ .)

Description

Note that the relation $xy^{-1} \in {\rm H}$ is compatible with right multiplication for any subgroup ${\rm H}$ : for any $a \in {\rm G}$ , $(xa)(ya)^{-1} = (xa)(a^{-1}y^{-1}) = xy^{-1} \in {\rm H}.$ On the other hand, if ${\rm H}$ is normal, then the relation must be compatible with left multiplication by any $a\in {\rm G}$ . This is true if and only $xy^{-1} \in {\rm H}$ implies $axy^{-1}a^{-1} = (ax)(ay)^{-1} \in {\rm H} .$ Since any element of ${\rm H}$ can be expressed as $xy^{-1}$ , the statement " ${\rm H}$ is normal in ${\rm G}$ " is equivalent to the following statement:

For all $a\in {\rm G}$ and $g\in {\rm H}$ , $aga^{-1} \in G$ ,

which is equivalent to both of the following statements:

For all $a \in {\rm G}$ , $a{\rm G}a^{-1} \subseteq {\rm G}$ ;
For all $a \in {\rm G}$ , $a {\rm G \subseteq G}a$ .

By symmetry, the last condition can be rewritten thus:

For all $a \in {\rm G}$ , $a {\rm G = G} a$ .

Examples

In an Abelian group, every subgroup is a normal subgroup.

Every group is a normal subgroup of itself. Similarly, the trivial group is a subgroup of every group.

Consider the smallest nonabelian group, $S_3$ (the symmetric group on three elements); call its generators $x$ and $y$ , with $x^3 = y^2 = (xy)^2 =e$ , the identity. It has two nontrivial subgroups, the one generated by $x$ (isomorphic to $\mathbb{Z}/3\mathbb{Z}$ and the one generated by $y$ (isomorphic to $\mathbb{Z}/2\mathbb{Z}$ ). Of these, the second is normal but the first is not.

If ${\rm G}$ and ${\rm G'}$ are groups, and $f: {\rm G \to G'}$ is a homomorphism of groups, then the inverse image of the identity of ${\rm G'}$ under $f$ , called the kernel of $f$ and denoted $\text{Ker}(f)$ , is a normal subgroup of ${\rm G}$ (see the proof of theorem 1 below). In fact, this is a characterization of normal subgroups, for if ${\rm H}$ is a normal subgroup of ${\rm G}$ , the kernel of the canonical homomorphism $f:{\rm G \to G/H}$ is ${\rm H}$ .

Note that if ${\rm H'}$ is a normal subgroup of ${\rm H}$ and ${\rm H}$ is a normal subgroup of ${\rm G}$ , ${\rm H'}$ is not necessarily a normal subgroup of ${\rm G}$ .

Group homomorphism theorems

Theorem 1. An equivalence relation $\mathcal{R}(x,y)$ on elements of a group ${\rm G}$ is compatible with the group law on ${\rm G}$ if and only if it is equivalent to a relation of the form $xy^{-1} \in {\rm H}$ , for some normal subgroup ${\rm H}$ of ${\rm G}$ .

Proof. One direction of the theorem follows from our definition, so we prove the other, namely, that any relation $\mathcal{R}(x,y)$ compatible with the group law on ${\rm G}$ is of the form $xy^{-1} \in {\rm H}$ , for a normal subgroup ${\rm H}$ .

To this end, let ${\rm H}$ be the set of elements equivalent to the identity, $e$ , under $\mathcal{R}$ . Evidently, if $x \equiv y \pmod{\mathcal{R}}$ , then $xy^{-1} \equiv e \pmod{\mathcal{R}}$ , so $xy^{-1} \in {\rm H}$ ; the converse holds as well, so $\mathcal{R}(x,y)$ is equivalent to the statement " $xy^{-1} \in {\rm H}$ ". Also, for any $x,y \in {\rm H}$ , $xy \equiv ee \equiv e \pmod{\mathcal{R}},$ so $xy \in {\rm H}$ . Thus ${\rm H}$ is closed under the group law on ${\rm G}$ , so ${\rm H}$ is a subgroup of ${\rm G}$ . Then by definition, ${\rm H}$ is a normal subgroup of ${\rm G}$ . $\blacksquare$

Theorem 2. Let ${\rm G}$ and ${\rm H}$ be two groups; let $f$ be a group homomorphism from ${\rm G}$ to ${\rm H}$ , and let ${\rm N}$ be the kernel of $f$ .

If ${\rm H'}$ is a subgroup of ${\rm H}$ , then the inverse image $f^{-1}({\rm H) = G'}$ of ${\rm H'}$ under ${\rm H}$ is a subgroup of ${\rm G}$ ; if ${\rm H'}$ is normal in ${\rm H}$ , then its inverse image is normal in ${\rm G}$ . Consequently, ${\rm N}$ is a normal subgroup of ${\rm G}$ , and of this inverse image. If $f$ is surjective, then $f({\rm G'}) = {\rm H'}$ , and $f$ induces an isomorphism from ${\rm G'/N}$ to ${\rm H'}$ .
If ${\rm G'}$ is a subgroup of ${\rm G}$ , then $f({\rm G'})$ is a subgroup of ${\rm H}$ ; if ${\rm G'}$ is normal in ${\rm G}$ , then $f({\rm G'})$ is normal in $f({\rm G})$ . In particular, if $f$ is surjective, then $f({\rm G'})$ is normal in ${\rm H}$ . The inverse image of $f({\rm G'})$ under $f$ is $\rm G'N = NG'$ .

Proof. For the first part, suppose $a,b$ are elements of ${\rm G'}$ . Then $f(ab) = f(a)f(b) \in {\rm H}$ , so $ab$ is an element of ${\rm G'}$ . Hence ${\rm G'}$ is a subgroup of ${\rm G}$ . If ${\rm H'}$ is a normal in ${\rm H}$ , then for all $a$ in ${\rm G}$ and all $b$ in ${\rm G'}$ , $f(a)f(b)f(a)^{-1} \in {\rm H'},$ so $aba^{-1} \in f^{-1}(\rm H') = G';$ thus ${\rm G'}$ is normal in ${\rm G}$ . Applying this result to the trivial subgroup of ${\rm H}$ , we prove that ${\rm N}$ is normal in ${\rm G}$ ; since the trivial subgroup of ${\rm H}$ is also a subgroup of ${\rm H'}$ , ${\rm N}$ is also a normal subgroup of ${\rm G'}$ . If $f$ is surjective, then by definition $f(\rm G') = H'$ . Also, if $a$ and $b$ are elements of ${\rm G'}$ which are congruent mod ${\rm N}$ , then $f(ab^{-1}) = f(e)$ , so $f(a) = f(b)$ . Thus $f$ induces an isomorphism from $\rm G'/N$ to $\rm H'$ which is evidently a homomorphism; hence, an isomorphism. This proves the first part of the theorem.

For the second part, suppose that $a,b$ are elements of ${\rm G'}$ . Then $f(a)f(b) = f(ab) \in f({\rm G'}) \subseteq f(\rm G) \subseteq H,$ so $f({\rm G'})$ is a subgroup of ${\rm H}$ and of $f({\rm G})$ . Suppose ${\rm G'}$ is normal in ${\rm G}$ . If $x$ is any element of ${\rm G}$ , then $f(x)f(a)f(x)^{-1} = f(xax^{-1}) \in f(\rm G') ,$ so $f( \rm G')$ is normal in $f(\rm G)$ . If $f$ is surjective, then $f(\rm G)= H$ , so $f({\rm G'})$ is normal in ${\rm H}$ .

Finally, suppose that $a$ is an element of ${\rm G}$ such that $f(a)$ is an element of $f({\rm G'}$ . Then for some $b \in \rm G'$ , $f(a) = f(b)$ . Hence $f(ab^{-1}) = f(a)f(b)^{-1} = f(e).$ Then $ab^{-1} = n$ , for some $n\in \rm N$ . Then $a= bn \in \rm G'N = NG'$ . This finishes the proof of the second part of the theorem. $\blacksquare$

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Normal subgroup

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Examples

Group homomorphism theorems

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