Normal Subgroups and Factor Groups

A subgroup \(H\) of a group \(G\) is \(\textbf{\textit{normal}}\) 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.

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 \(\textbf{\textit{facotr}}\) or \(\textbf{\textit{quotient group}}\) of \(G\) and \(N\).

:CUSTOM_ID: 89b66f4305be

\(Proof.\) Suppose \(H\) and \(H'\) is two normal subgroups of \(G\), \(N = H \cap H'\). \(gNg^{-1} = g(H \cap H')g^{-1} \subset H \cap H'\), so \(N\) is a normal group of \(G\). \(\blacksquare\)

\(Proof.\) \(G\) is abelian, so \[(aH)(bH) = abH = \{abh|h \in H\} = \{bah|h \in H\} = baH.\] Hence, \(G/H\) must also be abelian. \(\blacksquare\)

\(Solution.\) \(R_n\) is a normal subgroup of \(D_n\) for \(n \ge 3\), both \(R_n \) and \(D_n/R_n = \{R_n, sR_n\}\) are abelian, but \(D_n\) is not abelian. \(\blacksquare\)

\(Proof.\) If \(G\) is cyclic, then \(H\) must be cyclic and \(|H|\big | |G|\). Suppose \([G:H] = k\), then \(G/H = \{H, gH, g^2H, \cdots , g^{k-1}H\}\). Hence, \(G/H\) is cyclic too. \(\blacksquare\)

\(Solustion.\) \(R_n\) is a normal subgroup of \(D_n\) for \(n \ge 3\), both \(R_n \) and \(D_n/R_n = \{R_n, sR_n\}\) are cyclic, but \(D_n\) is not cyclic. \(\blacksquare\)

\(Proof.\) Suppose \([G:H] = 2\), then there are two cosets of G: \(\{H, gH\}\) where \(g \not \in H\). \(H\) must be a normal subgroup of \(G\) because for any \(g \in G\), if \(g \in H\), then \(gH = H = hg\); if \(g \not \in H\), then \(gH = Hg\), thus, \(H\) is a normal subgroup of \(G\).

For \(n \ge 3\), \(A_n\) is a subgroup of index \(2\) of \(S_n\), hence \(S_n\) is not simple. \(\blacksquare\)

\(Proof.\) For any \(g \in G\), \(gHg^{-1}\) is a subgroup of order \(k\) of \(G\). Now \(G\) has exactly one subgroup of order \(k\), so \(gHg^{-1} = H\) for any \(g \in G\). Thus, \(H\) is normal in \(G\). \(\blacksquare\)

\[C(g) = \{x \in G : xg = gx\}.\] Show that \(C(g)\) is a subgroup of \(G\). If \(g\) generates a normal subgroup of \(G\), prove that \(C(g)\) is normal in \(G\).

\(Proof.\) If \(x \in C(g)\), then \(xg = gx \rightarrow gx^{-1} = x^{-1}g\), thus \(x^{-1}\) is also in \(C(g)\). If \(x, y \in C(g)\), then \(xy^{-1}g = xgy^{-1} = gxy^{-1}\), so \(C(g)\) is a subgroup of \(G\).

For any \(h \in G\), \(x \in C(g)\),

\begin{align*} ghxh^{-1}g^{-1} &= ghx(gh)^{-1} \\ &= hg^kx(hg^k)^{-1} \\ &= hg^kxg^{-k}h^{-1} \\ &= hxh^{-1} \end{align*}

, so \(ghxh^{-1} = hxh^{-1}g\), thus \(hxh^{-1} \in C(g)\), which means \(hC(g)h^{-1} \subset C(g)\). Hence, \(C(g)\) is normal in \(G\). \(\blacksquare\)

\[Z(G) = \{x \in G : xg = gx \text{ for all } g \in G\}.\]

1. Calculate the center of \(S_3\).
2. Calculate the center of \(GL_2(\mathbb{R})\).
3. Show that the center of any group \(G\) is a normal subgroup of \(G\).
4. If \(G/Z(G)\) is cyclic, show that \(G\) is abelian.

\(Solution.\)

1. \(S_3 = \{id, (12), (13), (23), (123), (132)\}\), \(Z(S_3) = \{id\}\).
2. \(Z(GL_2(\mathbb{R})) = kE (k \not = 0)\) where \(E\) is the unit matrix.
3. It's obvious that \(gZ(G) = Z(G)g\), thus, \(Z(G)\) is normal in \(G\).
4. If \(G/Z(G)\) is cyclic, then \(\exists g \in G\), such that \(G/Z(G) = \langle gZ(G) \rangle\). For any \(a, b \in G\), \(a = g^kx\), \(b = g^ly\) where \(x\) and \(y\) are in \(Z(G)\). Thus, \(ab = g^kxg^ly = g^{k+l}xy = g^{k+l}yx = g^lyg^kx = ba\), which means \(G\) is abelian.

1. Show that \(G'\) is a normal subgroup of \(G\).
2. Let \(N\) be a normal subgroup of \(G\). Prove that \(G/N\) is abelian if and only if \(N\) contains the commutator subgroup of \(G\).

\(Proof.\)

1. \begin{align*} & g(aba^{-1}b^{-1})^kg^{-1} \\ = & (gaba^{-1}b^{-1}g^{-1})^k \\ = & (gag^{-1}gbg^{-1}ga^{-1}g^{-1}gb^{-1}g^{-1})^k \\ = & ((gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1})^k \end{align*}

   It's a product of a commutator, so \(g(aba^{-1}b^{-1})g^{-1} \in \langle aba^{-1}b^{-1} \rangle\) for any \(g \in G\). Thus \(G'\) is a normal subgroup of \(G\).

2. For any \(a,b \in G\),

   \begin{align*} & NaNb = NbNa \\ \Leftrightarrow & Nab = Nba \\ \Leftrightarrow & \exists n \in N, ab = nba \\ \Leftrightarrow & n = aba^{-1}b^{-1} \in N \qed \end{align*}