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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.

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Imibuzo abantu bebuza

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.

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IiNkqubo Abstract Algebra

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