Normal Subgroups and Quotient Groups

1. Definition of Normal Subgroups

In group theory, a subgroup $H$ of a group $G$ is called a normal subgroup if it is invariant under conjugation by any element of $G$. This means that for every element $g$ in $G$ and every element $h$ in $H$, the element $ghg^{-1}$ is also in $H$.

Mathematically, a subgroup $H \leq G$ is normal in $G$, denoted by $H \triangleleft G$, if and only if for all $g \in G$, $gHg^{-1} = H$. The set $gHg^{-1}$ is defined as $\{ghg^{-1} | h \in H\}$.

An equivalent condition for a subgroup $H$ to be normal is that its left cosets and right cosets are identical. That is, for every $g \in G$, the left coset $gH$ is equal to the right coset $Hg$. Remember that $gH = \{gh | h \in H\}$ and $Hg = \{hg | h \in H\}$.

Another way to define a normal subgroup is through the kernel of a homomorphism. The kernel of a group homomorphism $\phi: G \to G'$ is always a normal subgroup of $G$. Conversely, every normal subgroup of $G$ is the kernel of some homomorphism from $G$.

Let's look at some examples to solidify this concept.

Examples of Normal Subgroups

1. Trivial Subgroups: In any group $G$, the subgroup containing only the identity element, $\{e\}$, is always a normal subgroup. Also, the group $G$ itself is always a normal subgroup of $G$.

2. Abelian Groups: In an abelian group $G$ (where the operation is commutative, i.e., $ab = ba$ for all $a, b \in G$), every subgroup is a normal subgroup. This is because for any $g \in G$ and $h \in H$, $ghg^{-1} = (gh)g^{-1} = (hg)g^{-1} = h(gg^{-1}) = he = h$. Since $h \in H$, $ghg^{-1}$ is always in $H$.

3. Alternating Group $A_n$: The alternating group $A_n$, which consists of all even permutations in the symmetric group $S_n$, is a normal subgroup of $S_n$ for $n \ge 2$.

4. Center of a Group: The center of a group $G$, denoted by $Z(G) = \{z \in G | zg = gz \text{ for all } g \in G\}$, is always a normal subgroup of $G$. For any $g \in G$ and $z \in Z(G)$, $gzg^{-1} = (zg)g^{-1} = z(gg^{-1}) = ze = z$. Since $z \in Z(G)$, $gzg^{-1}$ is in $Z(G)$.

5. Non-Normal Subgroups: Consider the symmetric group $S_3$, the group of permutations of $\{1, 2, 3\}$. Let $H = \{e, (12)\}$ be a subgroup of $S_3$. Let $g = (13) \in S_3$. Then $gHg^{-1} = (13)\{e, (12)\}(13)^{-1} = \{(13)e(13)^{-1}, (13)(12)(13)^{-1}\}$. We know $(13)e(13)^{-1} = e$. Also, $(13)(12)(13)^{-1} = (13)(12)(13) = (132)$. The subgroup $H$ is $\{e, (12)\}$. The coset $gHg^{-1}$ is $\{e, (132)\}$. Since $\{e, (12)\} \neq \{e, (132)\}$, $H$ is not a normal subgroup of $S_3$.

2. Properties of Normal Subgroups

Normal subgroups have several important properties that make them fundamental in understanding group structure.

• The intersection of any collection of normal subgroups of $G$ is also a normal subgroup of $G$.
• The image of a normal subgroup under a group homomorphism is not necessarily normal in the image group, but the preimage of a normal subgroup under a homomorphism is always normal in the domain group.
• If $H \triangleleft G$ and $K \triangleleft G$, then $HK = \{hk | h \in H, k \in K\}$ is a subgroup of $G$. Furthermore, if $G$ is finite, then $|HK| = \frac{|H||K|}{|H \cap K|}$.
• If $H \triangleleft G$ and $K \triangleleft G$, then $H \cap K \triangleleft G$. Also, $H \cap K \triangleleft H$ and $H \cap K \triangleleft K$.
• If $H \triangleleft G$ and $K$ is any subgroup of $G$, then $H \cap K$ is a normal subgroup of $K$.

Let's prove one of these properties to see how it works.

Proof: Intersection of Normal Subgroups

Let $H_1$ and $H_2$ be two normal subgroups of $G$. We want to show that $H = H_1 \cap H_2$ is a normal subgroup of $G$.

First, we need to show that $H$ is a subgroup.

• Identity: Since $H_1$ and $H_2$ are subgroups, they both contain the identity element $e$. Thus, $e \in H_1 \cap H_2 = H$.
• Closure: Let $a, b \in H$. Then $a, b \in H_1$ and $a, b \in H_2$. Since $H_1$ and $H_2$ are subgroups, $ab \in H_1$ and $ab \in H_2$. Therefore, $ab \in H_1 \cap H_2 = H$.
• Inverses: Let $a \in H$. Then $a \in H_1$ and $a \in H_2$. Since $H_1$ and $H_2$ are subgroups, $a^{-1} \in H_1$ and $a^{-1} \in H_2$. Therefore, $a^{-1} \in H_1 \cap H_2 = H$.

So, $H$ is a subgroup of $G$.

Now, we need to show that $H$ is normal in $G$. Let $g \in G$ and $x \in H$. We need to show that $gxg^{-1} \in H$.

Since $x \in H$, we have $x \in H_1$ and $x \in H_2$. Because $H_1$ is a normal subgroup of $G$, for any $g \in G$ and $x \in H_1$, we have $gxg^{-1} \in H_1$. Similarly, because $H_2$ is a normal subgroup of $G$, for any $g \in G$ and $x \in H_2$, we have $gxg^{-1} \in H_2$.

Since $gxg^{-1}$ is in both $H_1$ and $H_2$, it must be in their intersection. Therefore, $gxg^{-1} \in H_1 \cap H_2 = H$.

This shows that $H = H_1 \cap H_2$ is a normal subgroup of $G$.

3. Definition of Quotient Groups

Normal subgroups are crucial because they allow us to form new groups called quotient groups (or factor groups). A quotient group is formed by taking a group $G$ and a normal subgroup $N$ of $G$, and considering the set of all left (or right) cosets of $N$ in $G$.

Let $G$ be a group and $N$ be a normal subgroup of $G$ ($N \triangleleft G$). The set of left cosets of $N$ in $G$ is denoted by $G/N = \{aN | a \in G\}$.

We can define a binary operation on the set $G/N$. For any two cosets $aN$ and $bN$ in $G/N$, we define their product as: $(aN)(bN) = (ab)N$.

We need to verify that this operation is well-defined. This means that if we choose different representatives for the cosets, the resulting coset is the same. Suppose $aN = a'N$ and $bN = b'N$ for some $a, a', b, b' \in G$. We need to show that $(ab)N = (a'b')N$.

Since $aN = a'N$, we know that $a' = an_1$ for some $n_1 \in N$. Since $bN = b'N$, we know that $b' = bn_2$ for some $n_2 \in N$. Then, $a'b' = (an_1)(bn_2) = a(n_1b)n_2$.

Now, we use the fact that $N$ is a normal subgroup. This implies that for any $x \in G$ and $y \in N$, $xyx^{-1} \in N$. In particular, for $b \in G$ and $n_1 \in N$, $bn_1b^{-1} \in N$. This means $bn_1 = (bn_1b^{-1})b$, and since $bn_1b^{-1} \in N$, let's call it $n_3$. So, $bn_1 = n_3b$ for some $n_3 \in N$.

Substituting this back into the expression for $a'b'$: $a'b' = a(n_3b)n_2 = (an_3)bn_2$. Since $a \in G$ and $n_3 \in N$, $an_3a^{-1} \in N$. Let $an_3a^{-1} = n_4 \in N$. Then $an_3 = n_4a$. So, $a'b' = (n_4a)bn_2 = n_4(ab)n_2$. This doesn't look like it directly leads to $(ab)N$. Let's use the property that $a' \in aN$ and $b' \in bN$.

A simpler way to show that the operation is well-defined: Suppose $aN = a'N$ and $bN = b'N$. This means $a' = ax$ for some $x \in N$ and $b' = by$ for some $y \in N$. Then $a'b' = (ax)(by) = a(xb)y$. Since $N$ is normal, $xb \in Nb$. And since $N$ is normal, $Nb=bN$, so $xb = bz$ for some $z \in N$. Thus, $a'b' = a(bz)y = (ab)zy$. Since $z \in N$ and $y \in N$, and $N$ is a subgroup, $zy \in N$. Therefore, $a'b' = (ab)w$ for some $w \in N$. This implies that $a'b'N = (ab)N$. The operation is indeed well-defined.

With this well-defined operation, the set $G/N$ forms a group. This group is called the quotient group of $G$ by $N$. The identity element of $G/N$ is the coset $eN = N$. The inverse of a coset $aN$ is $(aN)^{-1} = a^{-1}N$.

The order of the quotient group $G/N$ is the number of distinct cosets of $N$ in $G$, which is given by the index of $N$ in $G$, denoted by $[G:N]$. If $G$ is finite, then $|G/N| = |G|/|N|$.

Examples of Quotient Groups

1. Integers Modulo n: Consider the group of integers under addition, $(\mathbb{Z}, +)$. The set of even integers, $2\mathbb{Z} = \{..., -4, -2, 0, 2, 4, ...\}$, forms a subgroup of $\mathbb{Z}$. Since $\mathbb{Z}$ is abelian, $2\mathbb{Z}$ is a normal subgroup. The quotient group $\mathbb{Z}/2\mathbb{Z}$ consists of two cosets:

• $0 + 2\mathbb{Z} = \{..., -2, 0, 2, ...\}$ (the even integers)
• $1 + 2\mathbb{Z} = \{..., -1, 1, 3, ...\}$ (the odd integers)

The operation is addition of cosets: $(a+2\mathbb{Z}) + (b+2\mathbb{Z}) = (a+b)+2\mathbb{Z}$. If we add two even integers, we get an even integer (even coset + even coset = even coset). If we add an even and an odd integer, we get an odd integer (even coset + odd coset = odd coset). If we add two odd integers, we get an even integer (odd coset + odd coset = even coset). This quotient group $\mathbb{Z}/2\mathbb{Z}$ is isomorphic to the cyclic group of order 2, $C_2$ or $\mathbb{Z}_2$. In general, for any integer $n > 0$, $\mathbb{Z}/n\mathbb{Z}$ is a cyclic group of order $n$, denoted $\mathbb{Z}_n$.

2. Cyclic Groups: Let $G = \langle a \rangle$ be a cyclic group of order $n$. Any subgroup of a cyclic group is cyclic and is generated by some element $a^d$, where $d$ divides $n$. If $H = \langle a^d \rangle$, then $H$ is a normal subgroup of $G$ (since $G$ is abelian). The quotient group $G/H$ is isomorphic to the cyclic group $\mathbb{Z}_{n/d}$. For example, consider $G = \mathbb{Z}_6 = \{0, 1, 2, 3, 4, 5\}$ under addition modulo 6. The subgroup $H = \{0, 3\}$ is generated by 3, so $H = \langle 3 \rangle$. Here $n=6$ and $d=2$. The order of $H$ is 2. The quotient group $\mathbb{Z}_6/H$ has order $|G|/|H| = 6/2 = 3$. The cosets are:

• $0+H = \{0, 3\}$
• $1+H = \{1, 4\}$
• $2+H = \{2, 5\}$

The quotient group is isomorphic to $\mathbb{Z}_3$.

3. Dihedral Group $D_4$: Let $G = D_4$, the dihedral group of order 8 (symmetries of a square). Let $R$ be the rotation by 90 degrees and $F$ be a flip. The elements are $\{e, R, R^2, R^3, F, RF, R^2F, R^3F\}$. The subgroup $H = \{e, R, R^2, R^3\}$ (rotations) is a normal subgroup of $D_4$. The quotient group $D_4/H$ has order $|D_4|/|H| = 8/4 = 2$. The two cosets are $H$ itself and $F H = \{F, FR, FR^2, FR^3\}$. This quotient group is isomorphic to $\mathbb{Z}_2$.

4. Symmetric Group $S_3$: As we saw earlier, the subgroup $H = \{e, (12)\}$ is not normal in $S_3$. The normal subgroups of $S_3$ are $\{e\}$, $A_3 = \{e, (123), (132)\}$, and $S_3$. Let's consider the quotient group $S_3/A_3$. The order of $A_3$ is 3. The order of $S_3$ is 6. So, the order of the quotient group is $6/3 = 2$. The cosets are:

• $A_3 = \{e, (123), (132)\}$
• $(12)A_3 = \{(12)e, (12)(123), (12)(132)\} = \{(12), (13), (23)\}$

The quotient group $S_3/A_3$ is isomorphic to $\mathbb{Z}_2$.

4. The First Isomorphism Theorem for Groups

The concept of normal subgroups and quotient groups is central to understanding homomorphisms between groups. The First Isomorphism Theorem provides a fundamental link between these concepts.

Statement of the Theorem: Let $\phi: G \to G'$ be a group homomorphism. Let $K = \ker(\phi)$ be the kernel of $\phi$. Then $K$ is a normal subgroup of $G$. Furthermore, the quotient group $G/K$ is isomorphic to the image of $\phi$ in $G'$, denoted $\text{Im}(\phi)$. Mathematically: $G/\ker(\phi) \cong \text{Im}(\phi)$.

This theorem is extremely powerful. It tells us that every homomorphic image of a group $G$ is isomorphic to a quotient group of $G$ by some normal subgroup. The "some normal subgroup" is precisely the kernel of the homomorphism.

Proof Sketch: We have already established that the kernel $K$ of a homomorphism $\phi$ is a normal subgroup of $G$. We need to define an isomorphism $\psi: G/K \to \text{Im}(\phi)$. Let's define $\psi$ on a coset $aK \in G/K$ as $\psi(aK) = \phi(a)$. First, we must show that $\psi$ is well-defined. If $aK = bK$, then $b = ak$ for some $k \in K$. Applying $\phi$, we get $\phi(b) = \phi(ak) = \phi(a)\phi(k)$. Since $k \in K = \ker(\phi)$, we have $\phi(k) = e'$, where $e'$ is the identity in $G'$. Thus, $\phi(b) = \phi(a)e' = \phi(a)$. So, $\psi(bK) = \phi(b) = \phi(a) = \psi(aK)$. The map $\psi$ is well-defined.

Next, we show that $\psi$ is a homomorphism. $\psi((aK)(bK)) = \psi((ab)K) = \phi(ab)$. Since $\phi$ is a homomorphism, $\phi(ab) = \phi(a)\phi(b)$. And $\phi(a)\phi(b) = \psi(aK)\psi(bK)$. So, $\psi((aK)(bK)) = \psi(aK)\psi(bK)$, which means $\psi$ is a homomorphism.

Now, we show that $\psi$ is surjective onto $\text{Im}(\phi)$. By definition, $\text{Im}(\phi) = \{\phi(g) | g \in G\}$. For any element $y \in \text{Im}(\phi)$, $y = \phi(a)$ for some $a \in G$. Then $y = \phi(a) = \psi(aK)$. So, every element in the image of $\phi$ is hit by $\psi$.

Finally, we show that $\psi$ is injective. We need to show that if $\psi(aK) = e'$, then $aK$ is the identity element in $G/K$, which is $K$. If $\psi(aK) = e'$, then $\phi(a) = e'$. By the definition of the kernel, if $\phi(a) = e'$, then $a \in K$. If $a \in K$, then the coset $aK$ is equal to $K$ (since $aK = ak'e'$ for any $k' \in K$, and $aK=K$). Thus, $\psi$ is injective.

Since $\psi$ is a well-defined, surjective, injective homomorphism, it is an isomorphism. Therefore, $G/K \cong \text{Im}(\phi)$.

Applications of the First Isomorphism Theorem

The First Isomorphism Theorem is a cornerstone of abstract algebra and has numerous applications:

• Understanding Cyclic Groups: It proves that any cyclic group is isomorphic to either $\mathbb{Z}$ (if infinite) or $\mathbb{Z}_n$ for some $n$ (if finite). Consider the map $\phi: \mathbb{Z} \to \mathbb{Z}_n$ defined by $\phi(k) = k \pmod n$. This is a surjective homomorphism. The kernel is $\{k \in \mathbb{Z} | k \pmod n = 0\} = n\mathbb{Z}$. By the theorem, $\mathbb{Z}/n\mathbb{Z} \cong \mathbb{Z}_n$.
• Classifying Groups: It helps in classifying finite simple groups and understanding the structure of more complex groups by breaking them down into simpler quotient groups.
• Proving other Isomorphism Theorems: It is used as a basis for proving the other isomorphism theorems (Second, Third, and Fourth Isomorphism Theorems).

5. Criteria for Normal Subgroups

While the definition $gHg^{-1} = H$ is fundamental, it can sometimes be cumbersome to check. Here are some equivalent criteria that are often easier to use.

Let's illustrate condition 4 with an example.

Example using Condition 4:

Consider the group $G = D_3$ (symmetries of an equilateral triangle), which has order 6. Let $H = \{e, r, r^2\}$ be the subgroup of rotations, where $r$ is rotation by 120 degrees. We know $H \triangleleft D_3$ because $D_3$ has only one subgroup of order 3, and subgroups of odd order are normal. Let's verify using $gH = Hg$ for all $g \in D_3$. The elements of $D_3$ are $\{e, r, r^2, s, sr, sr^2\}$, where $s$ is a flip. We need to check for $g=s$. $sH = \{se, sr, sr^2\} = \{s, sr, sr^2\}$. $Hs = \{es, rs, r^2s\}$. We know that in $D_3$, $rs = sr^2$ and $r^2s = sr$. So, $Hs = \{s, sr^2, sr\}$. Thus, $sH = Hs$. This confirms that $H$ is normal.

Consider the subgroup $K = \{e, s\}$. Let $g = r$. $rK = \{re, rs\} = \{r, rs\}$. $Kr = \{er, sr\} = \{r, sr\}$. Since $rs \neq sr$ in $D_3$ (specifically, $rs=sr^2$), $rK \neq Kr$. Therefore, $K$ is not a normal subgroup of $D_3$.

6. Summary of Key Concepts

Normal subgroups and quotient groups are fundamental building blocks in group theory.

• A normal subgroup $N$ of $G$ is a subgroup that is closed under conjugation: $gNg^{-1} = N$ for all $g \in G$.
• Normal subgroups are precisely the kernels of group homomorphisms.
• Quotient groups $G/N$ are formed from the set of cosets of a normal subgroup $N$, with a well-defined group operation.
• The order of a quotient group $G/N$ is $|G|/|N|$ (if $G$ is finite).
• The First Isomorphism Theorem ($G/\ker(\phi) \cong \text{Im}(\phi)$) connects homomorphisms, kernels, normal subgroups, and quotient groups.
• Several equivalent criteria exist for checking if a subgroup is normal, often simplifying the verification process.

Understanding these concepts is crucial for further study in abstract algebra, including field theory, ring theory, and Galois theory.