maths.free › Abstract Algebra › Normal Subgroups and Factor Groups › Factor Groups and Normal Subgroups

Factor Groups and Normal Subgroups

A subgroup H of a group G is normal in G if gH = Hg for all g \in G. That is, a normal subgroup of a group G is one in which the right and left cosets are precisely the same. Example Let G be an abelian group.

Normal Subgroups

A subgroup \(H\) of a group \(G\) is normal in G if \(gH = Hg\) for all \(g \in G\). That is, a normal subgroup of a group \(G\) is one in which the right and left cosets are precisely the same.

Example

Let \(G\) be an abelian group. Every subgroup \(H\) of \(G\) is a normal subgroup. Since \(gh = hg\) for all \(g \in G\) and \(h \in H\), it will always be the case that \(gH = Hg\).

Example

Let \(H\) be the subgroup of \(S_3\) consisting of elements \((1)\) and \((12)\). Since \[\begin{aligned}\end{aligned}\], \(H\) cannot be a normal subgroup of \(S_3\). However, the subgroup \(N\), consisting of the permutations \((1)\), \((1 \, 2 \, 3)\), and \((1 \, 3 \, 2)\), is normal since the cosets of \(N\) are \[\begin{aligned}N = \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \} \\ (1 \, 2) N = N (1 \, 2) = \{ (1 \, 2), (1 \, 3), (2 \, 3) \}\end{aligned}\].

The following theorem is fundamental to our understanding of normal subgroups.

Theorem

Let \(G\) be a group and \(N\) be a subgroup of \(G\). Then the following statements are equivalent.

The subgroup \(N\) is normal in \(G\).
For all \(g \in G\), \(gNg^{-1} \subset N\).
For all \(g \in G\), \(gNg^{-1} = N\).

Proof

(1) \(\Rightarrow\) (2). Since \(N\) is normal in \(G\), \(gN = Ng\) for all \(g \in G\). Hence, for a given \(g \in G\) and \(n \in N\), there exists an \(n'\) in \(N\) such that \(g n = n' g\). Therefore, \(gng^{-1} = n' \in N\) or \(gNg^{-1} \subset N\).

(2) \(\Rightarrow\) (3). Let \(g \in G\). Since \(gNg^{-1} \subset N\), we need only show \(N \subset gNg^{-1}\). For \(n \in N\), \(g^{-1}ng=g^{-1}n(g^{-1})^{-1} \in N\). Hence, \(g^{-1}ng = n'\) for some \(n' \in N\). Therefore, \(n = g n' g^{-1}\) is in \(g N g^{-1}\).

(3) \(\Rightarrow\) (1). Suppose that \(gNg^{-1} = N\) for all \(g \in G\). Then for any \(n \in N\) there exists an \(n' \in N\) such that \(gng^{-1} = n'\). Consequently, \(gn = n' g\) or \(gN \subset Ng\). Similarly, \(Ng \subset gN\).

Factor Groups

If \(N\) is a normal subgroup of a group \(G\), then the cosets of \(N\) in \(G\) form a group \(G/N\) under the operation \((aN) (bN) = abN\). This group is called the factor or quotient group of \(G\) and \(N\). \(G/N\) factor group of \(G\) mod \(N\) Our first task is to prove that \(G/N\) is indeed a group.

Theorem

Let \(N\) be a normal subgroup of a group \(G\). The cosets of \(N\) in \(G\) form a group \(G/N\) of order \([G:N]\).

Proof

The group operation on \(G/N\) is \((a N ) (b N)= a b N\). This operation must be shown to be well-defined; that is, group multiplication must be independent of the choice of coset representative. Let \(aN = bN\) and \(cN = dN\). We must show that \[\begin{aligned}\end{aligned}\]. Then \(a = b n_1\) and \(c = d n_2\) for some \(n_1\) and \(n_2\) in \(N\). Hence, \[\begin{aligned}acN & = b n_1 d n_2 N \\ & = b n_1 d N \\ & = b n_1 N d \\ & = b N d \\ & = b d N\end{aligned}\]. The remainder of the theorem is easy: \(eN = N\) is the identity and \(g^{-1} N\) is the inverse of \(gN\). The order of \(G/N\) is, of course, the number of cosets of \(N\) in \(G\).

It is very important to remember that the elements in a factor group are sets of elements in the original group.

Example

Consider the normal subgroup of \(S_3\), \(N = \{ (1), (1 \, 2 \, 3), (1 \, 3 \, 2) \}\). The cosets of \(N\) in \(S_3\) are \(N\) and \((12) N\). The factor group \(S_3 / N\) has the following multiplication table.

\[\begin{aligned}\end{aligned}\]

This group is isomorphic to \({\mathbb Z}_2\). At first, multiplying cosets seems both complicated and strange; however, notice that \(S_3 / N\) is a smaller group. The factor group displays a certain amount of information about \(S_3\). Actually, \(N = A_3\), the group of even permutations, and \((1 \, 2) N = \{ (1 \, 2), (1 \, 3), (2 \, 3) \}\) is the set of odd permutations. The information captured in \(G/N\) is parity; that is, multiplying two even or two odd permutations results in an even permutation, whereas multiplying an odd permutation by an even permutation yields an odd permutation.

Condensed: the full section is in Judson, Abstract Algebra: Theory and Applications.

Jetzt du Kein Rechner löst diese Frage, aber ihre Teile sind berechenbar. Probieren Sie unten eine aus oder geben Sie Ihre eigene ein.

Behalten Sie Ihre eigenen Lösungswege

Ein kostenloses Konto fügt Notizen zu jeder Lektion, eine Aufzeichnung von dem, was Sie beendet haben, Ihre gelösten Probleme an einem Ort, und ein Tutor können Sie über diese Seite fragen. Die Mathematik selbst ist offen für alle, angemeldet oder nicht.

Registrieren Anmelden

Hier verwendete Symbole

Tippen Sie auf jedes Symbol für die vollständige Definition, ein Bild und was jeder Buchstabe darin bedeutet.

Fragen, die die Leute stellen

What is a group, in plain words?

A set with one operation that is associative, has an identity, and lets every element be undone. Symmetries of any object form a group. That is where the idea came from.

What is the difference between a ring and a field?

A ring has addition and multiplication that behave like the integers (you cannot always divide); a field is a ring where every non-zero element has a reciprocal, like the rationals or the reals.

Teile dieser Seite sind adaptiert aus Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Hier gekürzt und neu erklärt; Fehler gehen auf uns.

Mehr in Abstract Algebra

Groups Subgroups, cosets and Lagrange's theorem Cyclic groups and permutation groups Homomorphisms, normal subgroups and quotient groups Rings and fields Galois theory: why the quintic has no formula