Normal subgroup

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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 $n$.)


An equivalent definition of normal subgroups is this-

$N$ is said to be a normal subgroup of a group $G$ if $aNa^{-1}=N$.Note that this means $aN=Na$ but it does not imply that for every $n\in N$, $an=na$.

Description

From the characterizations of relations compatible with left and right translation (see the article on cosets), a subgroup $\rm H$ is normal if and only if $x^{-1}y \in \rm H$ is equivalent to $xy^{-1} \in \rm H$, which is in turn true if and only if $x^{-1}y \in \rm H$ implies $xy^{-1} \in \rm H$, which is in turn equivalent to its converse (by replacing $x$, $y$ with $x^{-1}$, $y^{-1}$).

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 $g\in {\rm H}$, $aga^{-1} \in H$,

which is equivalent to both of the following statements:

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

By symmetry, the last condition can be rewritten thus:

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

Equivalently, one can say that a normal subgroup is one that is stable under all inner automorphisms.

The intersection of a family $({\rm G}_i)$ of normal subgroups of a group $\rm G$ is a normal subgroup of $\rm G$. For $xy^{-1} \in {{\rm G}_i}$ (for each $i$) implies $x^{-1}y \in {\rm G}_i$ (for each $i$); hence $xy^{-1} \in \bigcap_i {\rm G}_i$ implies $x^{-1}y \in \bigcap_i {\rm G}_i$.

Examples

In an Abelian group, every subgroup is a normal subgroup. More generally, the center of every group is a normal subgroup of that group.

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 ${\rm G}$.

Every characteristic subgroup of $G$ is a normal subgroup of $G$.

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. $\blacksquare$

Corollary 3. Let $\rm G$ and ${\rm H}$ be groups; let ${\rm G'}$ be a subgroup of ${\rm G}$ and ${\rm L}$ a normal subgroup of ${\rm G'}$. Let $f$ be a group homomorphism from ${\rm G}$ to ${\rm H}$, with ${\rm N}$ the kernel of $f$. Then ${\rm LN}$ is a normal subgroup of ${\rm G'N}$, $\rm L \cdot (G' \cap N)$ is a normal subgroup of $\rm G'$, and $f({\rm L})$ is a normal subgroup of $f({\rm G'})$; furthermore, the quotient groups $\rm G'N/LN$, $\rm G'/L\cdot (G' \cap N)$, and $f({\rm G'})/f({\rm L})$ are isomorphic.

Proof. By theorem 2, $f({\rm L})$ is normal in $f({\rm G'})$. Let $\iota$ be the canonical homomorphism of $f({\rm G'})$ into $f({\rm G'})/f({\rm L})$; let $f'$ be the restriction of $f$ to $\rm G'N$, and let $g = \iota \circ f'$. Then $g$ is a surjective homomorphism from ${\rm G'N}$ to $f({\rm G'})/f({\rm L})$, and its kernel is ${\rm LN}$. Furthermore, $g$ induces a surjective homomorphism from ${\rm G'}$ to $f({\rm G'})/f({\rm L})$; the kernel of this homomorphism is ${\rm L \cdot (G' \cap N)}$. The corollary then follows from theorem 2. $\blacksquare$

For the following three corollaries, ${\rm G}$ will denote a group, and ${\rm N}$ a normal subgroup of $\rm G$, and $\lambda$ the canonical homomorphism from $\rm G$ to $\rm G/N$.

Corollary 4. The mapping $f: \rm G' \mapsto G'/N$ is a bijection from the set of subgroups of $\rm G$ that contain $\rm N$ to the set of subgroups of $\rm G'/N$.

Proof. Evidently, if $\rm G'$ is a subgroup of $\rm G$ containing $\rm N$, then $\rm G'/N$ is a subgroup of $\rm G/N$. If $\rm H$ is a subgroup of $\rm G/N$, then $\lambda^{-1}(\rm H)$ is a subgroup of $\rm G$ containing $\rm N$, so $f$ is surjective. Finally, since $\rm N$ is the kernel of $\lambda$, $(\lambda^{-1} \circ \lambda)(\rm G') = G'N = G'$, so $f$ is injective. $\blacksquare$

Corollary 5. Let $\rm G'$ be a subgroup of $\rm G$ containing $\rm N$. Then $\rm G'/N$ is normal in $\rm G/N$ if and only if $\rm G'$ is normal in $\rm G$; in this case, the groups $\rm G/G'$ and $\rm (G/N)/(G'/N)$ are isomorphic.

Proof. Note that $(\lambda^{-1} \circ \lambda)(\rm G') = G'N = G'$. Then by theorem 2, if $\lambda (\rm G')=G'/ N$ is normal in $\lambda(\rm G) = G / N$, then $\rm G'$ is normal in $\rm G$. Conversely, since $\lambda$ is surjective, if $\rm G'$ is normal in $\rm G$, then $\lambda(\rm G') = G'/N$ is normal in $\rm G'/N$. Now, suppose that $\rm G'$ is normal in $\rm G$. Let $f$ be the canonical homomorphism of $\rm G/N$ onto $\rm (G/N)/(G'/N)$. Evidently $f \circ \lambda$ is a surjective homomorphism from $\rm G$ to $\rm (G/N)/(G'/N)$, and the kernel of $f \circ \lambda$ is $\rm G'$. Then by theorem 2, $\rm G/G$ and $\rm (G/N)/(G'/N)$ are isomorphic. $\blacksquare$

Corollary 6. Let $\rm G'$ be a subgroup of $\rm G$. Then $\rm G'N$ is a subgroup of $\rm G$ of which $\rm N$ is a normal subgroup, and the groups $\rm G'/(G' \cap N)$ and $\rm G'N/N$ are isomorphic.

Proof. By theorem 2, $\rm G'N = NG'$ is the inverse image of the image of $\rm G'$ under $\lambda$; hence it is a subgroup of $\rm G$ in which $\rm N$ is evidently normal. Let $\iota$ be the canonical injection of $\rm G'$ into $\rm G'N$, and let $\lambda'$ be the restriction of $\lambda$ to $\rm G'N$. Then $\lambda' \circ \iota$ is a surjective homomorphism from $\rm G'$ to $\rm G'N/N$, and its kernel is $\rm (G' \cap N)$. Hence $\rm G'/(G' \cap N)$ and $\rm G'N/N$ are isomorphic, as desired. $\blacksquare$

See also

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