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.

Sasa Hakuna kifaa cha kupigia hesabu kinachoweza kufanya hesabu, lakini kifaa hicho kinaweza kutumiwa katika hesabu.

Keep your own working

Akaunti la bure laongezea mambo makuu juu ya kila somo, rekodi ya yale ambayo umemaliza, matatizo yako ya kutatua katika sehemu moja, na mtunzaji unayeweza kuuliza juu ya ukurasa huu.

Jisajili Login

Ishara zinazotumiwa hapa

Vaa alama yoyote ya ufafanuzi kamili, picha, na maana ya kila herufi.

Maswali ambayo watu huuliza

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.

Sehemu za ukurasa huu zimebadilishwa kutoka kwa Judson, Abstract Algebra: Theory and Applications (GFDL 1.3). Tumekamatwa na kukosolewa tena hapa; makosa ni yetu.

Mengi zaidi katika 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