Let \(N\) be a subgroup of \(G\). The following are equivalent:
- \(N\) is a normal subgroup of \(G\) (that is, \(xNx^{-1} = N\) for all \(x \in G\)).
- \(xNx^{-1} \subseteq N\) for all \(x \in G\).
- \(xN = Nx\) for all \(x \in G\).
Proof:
(1) \(\Rightarrow\) (2): This is immediate from the definition.
(2) \(\Rightarrow\) (1): Assume (2) holds. Let \(x \in G\). Then \(xNx^{-1} \subseteq N\). Replacing \(x\) by \(x^{-1}\) gives \(x^{-1}Nx \subseteq N\). Multiplying on the left by \(x\) and on the right by \(x^{-1}\) yields \(N = x(x^{-1}Nx)x^{-1} \subseteq xNx^{-1}\). Hence \(xNx^{-1} = N\).
(1) \(\Rightarrow\) (3): For any \(x \in G\), we have \(xNx^{-1} = N\). Multiplying on the right by \(x\) gives \(xN = Nx\).
(3) \(\Rightarrow\) (1): Assume \(xN = Nx\) for all \(x \in G\). Multiplying on the right by \(x^{-1}\) gives \(xNx^{-1} = N\). ∎
- Every subgroup of an abelian group is normal, because \(xhx^{-1} = hxx^{-1} = h \in H\).
- If \(H \le G\) has index \(2\), then \(H \trianglelefteq G\). To see this, note that the left cosets are \(H\) and \(G \setminus H\), and the right cosets are also \(H\) and \(G \setminus H\). For \(x \in H\), \(xH = H = Hx\). For \(x \notin H\), \(xH = G \setminus H = Hx\). Thus left and right cosets coincide.
- The center of a group \(G\), defined as \(Z(G) = \{z \in G : zx = xz \text{ for all } x \in G\}\), is a normal subgroup. For any \(x \in G\) and \(z \in Z(G)\), \(xzx^{-1} = zxx^{-1} = z \in Z(G)\).
Let \(N \trianglelefteq G\), and let \(G/N\) denote the set of left cosets of \(N\) in \(G\). The operation defined by
\[ (xN)(yN) = (xy)N \]is well defined, and makes \(G/N\) into a group.
Proof: We first show that the operation is well defined. Suppose \(xN = x'N\) and \(yN = y'N\). We must show that \((xy)N = (x'y')N\). Since \(xN = x'N\), there exists \(n_{1} \in N\) such that \(x' = xn_{1}\). Similarly, \(y' = yn_{2}\) for some \(n_{2} \in N\). Then
\[ x'y' = (xn_{1})(yn_{2}) = x(n_{1}y)n_{2}. \]Because \(N\) is normal, \(yN = Ny\), so \(n_{1}y = yn_{3}\) for some \(n_{3} \in N\). Thus
\[ x'y' = x(yn_{3})n_{2} = (xy)(n_{3}n_{2}). \]Since \(n_{3}n_{2} \in N\), we have \(x'y' \in (xy)N\), which means \((x'y')N = (xy)N\). Hence the operation is independent of the choice of coset representatives.
The group axioms follow immediately from those of \(G\):
- Associativity: \((xN \cdot yN) \cdot zN = (xy)N \cdot zN = ((xy)z)N = (x(yz))N = xN \cdot (yN \cdot zN)\).
- Identity: The coset \(eN = N\) acts as the identity, since \(xN \cdot N = (xe)N = xN\) and \(N \cdot xN = (ex)N = xN\).
- Inverses: The inverse of \(xN\) is \(x^{-1}N\), since \(xN \cdot x^{-1}N = (xx^{-1})N = eN = N\). ∎
The group \(G/N\) is called the quotient group (or factor group) of \(G\) by \(N\).
The map \(\pi : G \longrightarrow G/N\) defined by \(\pi(x) = xN\) is a surjective group homomorphism, called the canonical projection. The kernel of \(\pi\) is exactly \(N\).
Let \(\varphi : G \rightarrow H\) be a group homomorphism. The kernel of \(\varphi\) is
\[ \ker \varphi = \{g \in G : \varphi(g) = e_{H}\}, \]and the image of \(\varphi\) is
\[ \text{im } \varphi = \{\varphi(g) : g \in G\}. \]It is routine to verify that \(\ker \varphi\) is a normal subgroup of \(G\), and \(\text{im } \varphi\) is a subgroup of \(H\).
Let \(\varphi : G \longrightarrow H\) be a group homomorphism, and let \(K = \ker \varphi\). Then the map
\[ \bar{\varphi} : G/K \longrightarrow \text{im } \varphi, \quad \bar{\varphi}(xK) = \varphi(x) \]is a well-defined group isomorphism. Consequently, \(G/\ker \varphi \cong \text{im } \varphi\).
Proof: To see that \(\bar{\varphi}\) is well defined and injective, note that
\[ xK = yK \iff y^{-1}x \in K \iff \varphi(y^{-1}x) = e_{H} \iff \varphi(y)^{-1}\varphi(x) = e_{H} \iff \varphi(x) = \varphi(y). \]The first implication shows \(\bar{\varphi}\) is well defined; the reverse implication shows it is injective. It is surjective onto \(\text{im } \varphi\) by definition. Finally, it is a homomorphism:
\[ \bar{\varphi}((xK)(yK)) = \bar{\varphi}((xy)K) = \varphi(xy) = \varphi(x)\varphi(y) = \bar{\varphi}(xK)\bar{\varphi}(yK). \quad \blacksquare \]Let \(G\) be a group, \(H \le G\), and \(N \trianglelefteq G\). Then \(HN = \{hn : h \in H, n \in N\}\) is a subgroup of \(G\), \(H \cap N \trianglelefteq H\), and
\[ H / (H \cap N) \cong HN / N. \]Let \(G\) be a group, and let \(K \le N \le G\) with both \(K\) and \(N\) normal in \(G\). Then \(N/K\) is a normal subgroup of \(G/K\), and
\[ (G/K) / (N/K) \cong G/N. \]Let \(N \trianglelefteq G\). There is a bijection between the set of subgroups of \(G\) containing \(N\) and the set of subgroups of \(G/N\), given by
\[ H \longmapsto H/N. \]Furthermore, this correspondence preserves inclusions, indices, and normality: \(H \trianglelefteq G\) if and only if \(H/N \trianglelefteq G/N\).