1.6 Normal Subgroups

Definition 1.6.1. A subgroup \(K\) of a group \(G\) is called a normal subgroup of \(G\) (is normal in \(G\)) if for each element \(k\in K\), and \(g\in G\) we have \(gkg^{-1}\in K\).

Note. By this definition, then \(K\) is normal in \(G\) if for all \(g\in G\hspace {0.3cm} gk = kg\) where \(gk = kg \implies \, gk = k'g\) where \(k\, , \, k'\in K\) or \(k = k'\).
* \(K\) “commutes” as a set with elements of \(G\).

Example 1.6.2.

1.

The Cyclic group \(H = \langle (12)\rangle \) of \(S_3\) consisting of \((1)\,,\, (12)\) is not normal in \(S_3\). Because.
Let for instance \(\, g = (1\, 2\, 3)\, , \,\, g^{-1} = (3\, 2\, 1)\).
We have \(\, g(12)g^{-1} = (1\, 2\, 3)(1\,2)(3\,2\,1) = (2\,3) \not \in H\). Hence \(H\) is not normal in \(S_3\).

2.

However, the Cyclic subgroup \(\, \langle (1\, 2\, 3)\rangle \,\) of \(S_3\) is a normal subgroup. (Check this is exercise)

\(H = \langle (1\, 2\, 3)\rangle = \big \{(1\,2\,3)^n:\, 1\leq n\leq \begin {vmatrix} (1\, 2\, 3)\\ \end {vmatrix}\big \} = \big \{(1\,2\,3)^n:\, 1\leq n\leq 3\big \} = \big \{(1\,2\,3)\, ,\, (1\, 3\, 2)\, , \, (1)\big \}\)

(a): If \(\, g = (1)\) then \(g^{-1} = (1)\,\implies gkg^{-1} = k\in K\)
(b): If \(\, g = (1\, 2) \implies g^{-1} = (2\,1) \implies g(1\, 2\, 3)g^{-1} = (1\,2)(1\, 2\, 3)(2\,1) = (1\,3\, 2)\in K\)
(c): If \(\, g= (1\,3) \implies g^{-1} = (3\, 1) \implies g(1\, 2\, 3)g^{-1} = (1\,3)(1\,2\,3)(3\,1) = (1\,3\,2)\in H = K\)
(d): If \(\, g = (1\,2\,3) \, , \, g^{-1}(3\, 2\, 1) \implies g(1\,3\,2)g^{-1} = (1\, 2\, 3)(1\,3\,2) (3\,2\,1) = (1\,3\,2)\in K\)
(e): If \(\, g = (2\,3)\,,\, g^{-1} = (3\,2) \implies g(1\, 2\, 3)g^{-1} = (2\, 3)(1\,2\,3)(3\,2) = (1\,3\, 2)\in K\)
(f): If \(\, g = (1\, 3\,2) \, , \, g^{-1} = (2\, 3\, 1)\implies g(1)g^{-1} = (1\, 3\, 2) (1)(2\,3\,1) = (1)\in K \)

\(\therefore \, \, \forall \, g\in S_3\) and \(k\in K\,\, gkg^{-1}\in K\)

3.

Every subgroup \(H\) of a Abelian group \(G\) is normal in \(G\).

Proof. If \(G\) is an Abelian group, every subgroup \(H\) of \(G\) is normal in \(G\) since, if \(g\in G\), then \(gH = Hg\implies gHg^{-1} = H \implies gHg^{-1} \in H\). Let \(h\in H\), then \(ghg^{-1} \in H\,\, \forall \, h\in H\). Therefore, \(H\) is normal in G.
□

4.

Let \(H= SL(2\, ,\, \mathbb {R})\hspace {0.5cm} G = GL(2\, , \, \mathbb {R})\)

\(2 \implies \, 2\times 2\, \) matrices \(\hspace {1cm} R \implies \) entries from \(\mathbb {R}\).

Then \(H\) is a normal subgroup of \(G\). \begin {align*} G = \Big \{\begin {pmatrix} a & b\\ c & d\\ \end {pmatrix}\Big |\, a, \, b, \, c, \, d\in \mathbb {R}\, \hspace {0.3cm} \text {and}\hspace {0.3cm}\begin {vmatrix} a & b\\ c & d\\ \end {vmatrix} \neq 0\Big \}\\\\ H = \Big \{\begin {pmatrix} a & b\\ c & d\\ \end {pmatrix}\Big |\, a,\, b,\, c,\, d\in \mathbb {R},\hspace {0.3cm} \text {and}\hspace {0.3cm} \begin {vmatrix} a & b\\ c & d\\ \end {vmatrix} = 1\Big \} \end {align*}

Solution. Clearly, \(H\neq \emptyset \) because \(\begin {pmatrix} 1 & 0\\ 0 & 1\\ \end {pmatrix}\in H\). Now for any elements \(A, \, B\in H\), we want to show that \(AB^{-1}\in H\).

\begin {align*} det\big (AB^{-1}\big ) & = \det (A)\cdot \det (B^{-1}) = \det (A)\cdot \dfrac {1}{\det (B)}\\ & = 1\cdot 1/1 = 1\, \implies A,\, B^{-1}\in H \end {align*}

i.e \(H\) is a subgroup of \(G\).
Now we check if \(H\,\vartriangleleft \,G\) (if \(H\) is normal in \(G\)). i.e, we want to show that, for \(A\in H,\,\, B\in G,\,\, BAB^{-1}\in H\). Now \begin {align*} \det \big (BAB^{-1}\big ) & = \det B\cdot \det A\cdot \dfrac {1}{\det B} = \det (A) = 1 \end {align*}

Hence \(\, BAB^{-1}\, \) and \(\, H\lhd G\).

Exercise \(*\)
Show that

1.: \(SL (n\, ,\, \mathbb {R}) \lhd GL(n\, ,\, \mathbb {R})\)
2.: \(SL (n\, , \, \mathbb {C})\lhd GL(n\,,\, \mathbb {C})\)
3.: \(SL (2n\, , \, \mathbb {R})\lhd GL(2n\, , \, \mathbb {R})\)

\(4.\) For any \(n, \, A_n \lhd S_n\).

Proof. Indeed \(A_n < S_n\), we just need to show that it is normal in \(S_n\).
Let \(\alpha \in A_n \, ,\, \beta \in S_n\). We want to show that \(\beta \alpha \beta ^{-1}\in A_n\).
Now \(Sgn\big (\beta \alpha \beta ^{-1}\big ) = Sgn(\beta )\, Sgn(\alpha )\, Sgn(\beta ^{-1}) = Sgn(\beta )\, Sgn(\beta ^{-1}) = 1\) (since if \(\beta \) is even also \(\beta ^{-1}\) and if \(\beta \) is odd so is \(\beta ^{-1}\)). Thus \(\beta \alpha \beta ^{-1}\) is even \(\,\implies \beta \alpha \beta ^{-1}\in A_n\). Therefore \(A_n\lhd S_n\).

□

Definition 1.6.3. A group \(G\neq \{e\}\) is called a simple group if \(G\) has no normal subgroups except \(\{e\}\) and \(G\).

\(\bullet \) We now introduce some special subgroup which are normal in any group.

Definition 1.6.4. Let \(G\) be a group. Then center of \(G\) denoted as \(Z(G)\) is defined as \(Z(G) = \{z\in G:\, zg = gz\, , \, \) for all \(g\in G\}\).

Exercise 1.6.5. Show that the centre of a group \(G\) is a normal subgroup of \(G\).

Proof. Let \(\, z\in Z(G)\) and \(g\in G\) then \(zg = gz\) since \(Z(G)\) is the centre of \(G\). This implies that \(z = gzg^{-1}\in Z(G)\,\, \forall g\in G\). Therefore \(Z(G)\) is normal subgroup of \(G\).

□

Note. A group \(G\) is Abelian if and only if \(Z(G) = G\). On the other hand a group is said to centerless if \(Z(G) = \{e\}\).

Example 1.6.6.

1.

\(G = \Bigg \{\begin {pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end {pmatrix}:\, a,\, b,\, c\in \mathbb {R}\Bigg \}\) “The Heisenberg Group”.

Then \(Z(G) = \Bigg \{\begin {pmatrix} 1 & 0 & d\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end {pmatrix}:\, d\in \mathbb {R}\Bigg \}\)

To show this let \(A = G = \begin {pmatrix} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end {pmatrix}\, \) and \(\, B = \begin {pmatrix} 1 & 0 & d\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end {pmatrix}\,\) and show that \(\, AB = BA\).

2.

Let \(G\) be non-Abelian and simple (i.e \(G\) is not commutative and has no normal subgroup except \(\{e\}\) and \(G\)). Then \(Z(G) = \{e\}\).

Proof. by nemwine!!!
Let \(G\) be non-Abelian and simple. Then \(G\) has only two normal subgroups \(G\) and \(\{e\}\) since \(G\) is simple. Suppose \(a,\, b\in G\) then \(ab\neq ba\) since \(G\) is non-Abelian. However, since \(G\) is simple then \(\, \forall g\in G\, \exists \, e\in G\) such that \(ge = eg = ge\in G\). This implies that the only element of \(G\) that commutes with all other elements is the identity \((e)\). Thus \(\forall g\in G:\, g\neq e\in G\, ,\, g\not \in Z(G)\implies Z(G) = \{e\}\). Thus the center of one element only \((e)\).
□

3.

Let \(G = D_n - \) dihedral group on an \(n-g\) on of order \(2n\).
If \(n\) is odd, \(Z(G) = \{e\}\).
If \(n\) is even , \(Z(G) = \{e\, , \, r_{180}\}\).

Definition 1.6.7. Let \(x\, y\in G\), where \(G\) is a group. The elements \(x^{-1}y^{-1}xy\) in \(G\) is called the commutator. The elements \(x^{-1}y^{-1}xy\) is dented as \([x\, , \, y]\).

Definition 1.6.8. The subgroup of \(G\) generated by all commutators is called the commutator subgroup of \(G\). We denote this as \(G'\). It is sometimes known as the derived subgroup of \(G\).
(i.e \(G' = \big \{\langle x^{-1}y^{-1}xy\rangle \, x,\, y\in G\big \}\).

Exercise \(**\)

1.

Find \(G',\,\) if \(G = \mathbb {Z}^*_{12}\, - \) where \(\mathbb {Z}^*_{n} = \) multiplicative group \(\, \mathbb {Z}^*_{\rho } = \{x\in \mathbb {Z}_n:\, (x\, , \, n) = 1\}\).

2.

Find \(\, G',\, \) if \(G = S_3\)

e.g \(\, \mathbb {Z}^*_{6} = \{1\, ,\, 5\}\)

\(\mathbb {Z}^*_{12} = \{1,\, 5,\, 7,\, 11\}\)

\(-\, U(n) = \) group of units \(= \{x\in \mathbb {Z}_n:\, x^2 = 1\}\)

e.g \(U(8) = \{1,\, 3, \, 5\, 7\}\)

Definition 1.6.9. Let \(A\) be a subset of \(G\). The centraliser \(C(A)\) of \(A\) in \(G\) is defined as
\(C(A) = \{c\in G:\, ca = ac\,\,\forall a\in A\}\)

Lemma 1.6.10. \(C(A)\) is a normal subgroup of \(G\).

Proof. (Exercise)(By nemwine!!!!)
Let \(A\) be a group of \(G\) such that \(\, \forall a\in A\) and \(g\in G\), then \(ca = ac\). This is a normal subgroup of \(G\) since form \(ca = ac\) we have that \(a = cac^{-1}\in A\,\, \forall a\in A\). Thus \(A\lhd G\).

□

Lemma 1.6.11. If \(A\) is an Abelian group, then \(A\) is normal in \(C(A)\).

Proof. Let \(A\) be abelian and let \(C(A)\) be its centraliser, so every element of \(C(A)\) commutes with every element of \(A\). Take \(a\in A\) and \(c\in C(A)\). Then \[cac^{-1} = acc^{-1} = a \in A ,\] using that \(c\) commutes with \(a\). So \(cAc^{-1}\subseteq A\) for every \(c\in C(A)\), which is precisely the statement that \(A\) is normal in \(C(A)\). □

Definition 1.6.12. The normaliser \(N_G(A)\) of \(A\) in \(G\) is defined as; \[N_G(A) = \{n\in G:\, A_n = nA\} = \{n\in G:\, nA^{-1}n = A\}\]

Note;
\(N_G(A)\) is a subgroup of \(G\). If \(A\) is a subgroup, then \(N_G(A)\lhd G\). In the case where \(A\) is a subgroup of \(G\), then \(A\lhd G\) if and only if \(N_G(A) = G\).

Proposition 1.6.13.

(i): If \(H\) is a subgroup of index 2 in a group \(G\), then \(g^2\in H\,\,\, \forall g\in G\) or if \(H\) is of index 2 then it contains all the squares.
(ii): If \(H\) is a subgroup of index 2 in a group \(G\), then \(H\,\vartriangleleft \, G\).

Proof.

(i): Since \(H\) is of index 2, then \(\begin {vmatrix} G\\ \end {vmatrix} = 2\, \begin {vmatrix} H\\ \end {vmatrix}\). Hence there are exactly two cosets, namely \(H\) and \(a H\) for \(a\in H\) i.e \(G = H\cup aH\). Let \(g\in G\,\ni \, g\not \in H\), then \(g = ah\) for some \(h\in H\). If \(g^2\not \in H\) then, \(g^2 = ah'\) for some \(h'\in H\). Then \(g = g^{-1}g^2 = h^{-1}a^{-1}ah' = h^{-1}h'\in H\). This is a contradiction since it was assumed that \(g\not \in H\). Hence \(g^2\in H\).
(ii): To prove the second statement, it is sufficient to show that if \(h\in H\), \(\, \, ghg^{-1}\in H\,\,\, \forall g\in G\). \(H\) is of index 2 in \(G\) implies there are exactly 2 cosets namely \(H\) and \(aH\) for some \(a\not \in H\). Let \(g\in G\), then \(g\in H\) or \(g\in aH\). If \(g\in H\), then clearly \(ghg^{-1}\in H\) because \(H\) is a subgroup.
If \(ghg^{-1}\not \in H\), then, \(ghg^{-1} = ah'a^{-1}\in aH\) so that; \(ah'a^{-1} = ay\) for some \(y\in H\). Canceling \(a\), we have \(h'a^{-1} = y\). This implies that \(a = y^{-1}h'\in H\). This is a contradiction since \(a\not \in H\). Therefore, \(ghg^{-1}\in H\) so if \(h\in H\) then every. Conjugate of \(h\) also lies in \(H\). Hence \(H\vartriangleleft G\).

□

Definition 1.6.14. If \(H\) is a subgroup of a group \(G\), then the conjugate of it is a subgroup of \(G\) of the form; \(\, aHa^{-1} = \{aha^{-1}\, :\, h\in H\}\,\) where \(a\in G\).

Definition 1.6.15. If \(H\) and \(K\) are subgroups of a group \(G\), we say that \(H\) and \(K\) are conjugate in \(G\) if \(\, \exists \,\) an element \(g\in G\) such that \(H = gKg^{-1}\).

SOLUTIONS TO THE ASSIGNMENT

(i): Suppose \(H\) and \(K\) are subgroups of \(G\) with \(K\vartriangleleft G\). Then \(HK = \{hk\, |\, h\in H\, ,\, k\in K\}\). Clearly, \(HK\neq \emptyset \) since \(e\cdot e = e\in HK\) for some \(e\in H\) and \(e\in K\) since both \(H\) and \(K\) are subgroups. Now we must show that if \(a\) and \(b\) are in \(HK\) then \(ab^{-1}\in HK\). Let \(a = h_1k_1\) and \(b = h_2k_2\) for \(h_1\, , h_2\in H\) and \(k_1\, , k_2\in K\). Then \[ab^{-1} = h_1k_1(h_2k_2)^{-1} = h_1k_1k_2^{-1}h_2^{-1} = h_1ek_1k_2^{-1}h_2^{-1} = h_1h_2^{-1}h_2k_1k_2h_2^{-1}\] \(\implies h_1h_2^{-1}\in H\) and \(h_2k_1k_2^{-1}h_2^{-1}\in K\) since \(K\vartriangleleft G\).
Hence \(ab^{-1} = h_1k_1(h_2k_2)^{-1}\in HK\), so \(HK\) is a subgroup of \(G\).
(ii): Let \(g\in G\) and \(x\in HK\). Then \(x = hk\) for some \(h\in H\) and \(k\in K\). Thus
\(gxg^{-1} = g(hk)g^{-1} = \big (ghg^{-1}\big )\big (gkg^{-1}\big )\in HK\) (since \(ghg^{-1} \in H\) and \(gkg^{-1}\in K\)). This proves that \(HK\) is a normal subgroup of \(G\).

Solutions to Exercise “*”

(a): Let \(G = GL_{(n,\mathbb {R})}\) and \(N = SL_{(n,\mathbb {R})}\). We need to show that \(gxg^{-1}\in N\,\, \forall x\in N\) and \(g\in G\). This translates into the case of \(P\) invertible matrices. Any invertible matrix \(P\) and \(\det (A) = 1\), we have \(PAP^{-1} = 1\, \,\) since \[\det \big (PAP^{-1}\big ) = \det (P)\, \det (A)\, \dfrac {1}{\det (P)} = \det (A) = 1\] Thus \(\, \forall A\in N\) and \(P\) an \(n\times n\) invertible matrix in \(G\), we obtain \(PAP^{-1}\in N\). Hence \(SL_{(n,\mathbb {R})}\vartriangleleft GL_{(n,\mathbb {R})}\).
(b).

Exercise “**”

1.: \(\mathbb {Z}^*_{12} = \{x\in \mathbb {Z}\, :\, (x,12) = 1\}\)
\(\implies \, \mathbb {Z}^*_{12} = \{1\, , \, 5\, , \, 7\, , \, 11\}\,\) here, the identity is 1 implying (\(*\) is multiplication)
\(\implies \, \big (\mathbb {Z}^*_{12} = \big \{1\, , \, \frac {1}{5}\, , \, \frac {1}{7}\, , \, \frac {1}{11}\big \}\)

Thus, \(\, \forall x\in \mathbb {Z}^*_{12}\,\) we have \(xx^{-1} = e\cdot = 1\). Hence \(\, G' = \{1\}\) since
\(\, G' = \{\langle x^{-1}y^{-1}xy\rangle = \langle 1\rangle = 1\}\,\, \forall x\, ,\,y \in \mathbb {Z}^*_{12}\)
2.: \(G = S_3 = \{(1)\, , \, (12)\, , \, (13)\, , \, (23)\,, \, (123)\, , \, (132)\}\)
We know that \((1)\) is an identity \(\, \implies \, \forall x\in S_3\hspace {0.5cm} xe = ex = x\). To identify any other commutator element we try to pair the elements in \(S_3\).
\(\, (12)(13) = (132)\ne (123) = (13)(12)\). Hence \((12)\) nor \((13)\) is a commutator. However we have \(\, (21)(31)(12)(13) = (123)\,\) is also a commutator.
\((23)(123) = (13)(2)\neq (12)(3) = (123)(23)\). Hence neither \((123)\) nor \((23)\) is a commutator. However, \((32)(321)(23)(123) = (132)\) is a commutator.

Proof that \(132\) is a commutator \[(123)(132) = (1)(2)(3) = (1)(2)(3) = (132)(123)\] \[(1) = (1)\] Hence \(\, {[x,y]} = \{(1)\, , \, (123)\, , \, (132)\}\)

Now \(\, G' = \big \{\langle (1)\, , \, (123)\, , \, (132)\rangle \big \} = \big \{(1)\, ,\, (123)\, , \, (132)\big \}\)

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