A normal subgroup of a subgroup follows the standard definition of a normal subgroup, where the subgroup itself acts as the encompassing group. However, a critical aspect to understand is that normality is not a transitive relation: a normal subgroup of a normal subgroup of a larger group does not necessarily mean it's normal in the largest group.
What is a Normal Subgroup?
In group theory, a normal subgroup is a special type of subgroup that remains invariant under conjugation by elements of the larger group. Formally, a subgroup $N$ of a group $G$ is called a normal subgroup (denoted $N \triangleleft G$) if for every element $g \in G$ and every element $n \in N$, the element $gng^{-1}$ is also in $N$. This essentially means that conjugating any element of $N$ by any element from $G$ results in an element that stays within $N$. Normal subgroups are fundamental because they are precisely the kernels of group homomorphisms, playing a crucial role in constructing quotient groups.
Normal Subgroups Within Subgroups
When we refer to a "normal subgroup of a subgroup," we are applying the standard definition of normality in a nested context. Let $G$ be a group, and let $H$ be a subgroup of $G$ ($H \leq G$). Then, a subgroup $K$ of $H$ ($K \leq H$) is defined as a normal subgroup of $H$ (denoted $K \triangleleft H$) if:
- $K$ is a subgroup of $H$.
- For every element $h \in H$ and every element $k \in K$, the conjugate $hkh^{-1}$ is an element of $K$.
In this scenario, $H$ acts as the "parent group" for $K$, and the definition of normality applies directly to $K$ within the context of $H$.
Is Normality a Transitive Property?
This is a key concept where many find a common misconception. The property of being a normal subgroup is not transitive. This means that if you have a chain of subgroups, such as $K \triangleleft H$ (K is normal in H) and $H \triangleleft G$ (H is normal in G), it does not automatically imply that $K \triangleleft G$ (K is normal in G).
Why Normality is Not Transitive
The definition of normality is relative to its immediate parent group.
- The condition $K \triangleleft H$ means that $K$ is closed under conjugation by elements from $H$.
- The condition $H \triangleleft G$ means that $H$ is closed under conjugation by elements from $G$.
- For $K$ to be normal in $G$ ($K \triangleleft G$), it must be closed under conjugation by all elements from $G$.
The issue arises because an element $g \in G$ might not be in $H$. While $K$ is invariant under conjugation by elements of $H$, it might not be invariant under conjugation by elements of $G$ that are outside $H$.
Illustrative Example: The Dihedral Group of Order 8
The smallest group exhibiting this non-transitive phenomenon is the dihedral group of order 8 ($D_4$), which represents the symmetries of a square.
Let $G = D_4$. We can describe $D_4$ using two generators: a rotation $r$ (by 90 degrees) and a reflection $s$. The group elements are ${e, r, r^2, r^3, s, sr, sr^2, sr^3}$.
Consider the following chain of subgroups:
- Define $H$: Let $H = {e, r^2, s, sr^2}$. This subgroup is isomorphic to the Klein four-group and is normal in $D_4$ ($H \triangleleft D_4$).
- Verification: You can check that for any $g \in D_4$ and $h \in H$, $ghg^{-1} \in H$. For example, $r(s)r^{-1} = sr^2 \in H$.
- Define $K$: Let $K = {e, s}$. This subgroup is clearly a subgroup of $H$ ($K \leq H$).
- Check $K \triangleleft H$: Since $H$ is an abelian group (meaning its elements commute, e.g., $sr^2 = r^2s$), all its subgroups are normal. Thus, $K \triangleleft H$.
- Check $K \triangleleft G$ (The Test for Transitivity): Now, we need to see if $K$ is normal in the larger group $G=D_4$.
- Take an element $k = s \in K$.
- Take an element $g = r \in G$ (note that $r \notin H$).
- We compute the conjugate: $r s r^{-1}$. Using the relations of $D_4$ ($sr = r^{-1}s$), we get $r s r^{-1} = r(r^{-1}s) = s$.
- Let's re-evaluate. $rsr^{-1}=rsr^3$. From $sr=r^{-1}s$, we have $s=r^{-1}sr$. Then $rsr^{-1} = s$. This means it is normal for this element.
My example choice for $D_4$ earlier was correct, $r s r^{-1} = s r^2$.
Let's re-confirm the $D_4$ example carefully.
In $D_4$, $r^4=e, s^2=e, srs^{-1}=r^{-1}$.
Let $G = D_4$.
Let $H = {e, r^2, s, sr^2}$. This is normal in $D_4$. ($H \triangleleft D_4$).
Let $K = {e, s}$. This is a subgroup of $H$.
Is $K \triangleleft H$? Yes. $r^2 s (r^2)^{-1} = r^2 s r^2 = (r^2 s) r^2 = s r^{-2} r^2 = s$. So $K$ is normal in $H$.
Is $K \triangleleft G$?
Take $k = s \in K$. Take $g = r \in G$.
$r s r^{-1} = r s r^3$. Since $srs^{-1}=r^{-1}$, then $sr=r^{-1}s$.
So $r s r^{-1} = r(r^{-1}s)r^2 = s r^2$.
Is $s r^2 \in K = {e, s}$? No, because $r^2 \neq e$.
Therefore, $K$ is not normal in $G$.
This counterexample clearly demonstrates that normality is not a transitive property.
Characteristic Subgroups: A Transitive Alternative
While normality is not transitive, a related and stronger concept called a characteristic subgroup is transitive in a specific way. A subgroup $K$ of $G$ is called a characteristic subgroup (denoted $K \text{ char } G$) if it is invariant under all automorphisms of $G$.
A key property derived from this definition is that a characteristic subgroup of a normal subgroup is normal in the larger group. That is, if $K \text{ char } H$ and $H \triangleleft G$, then it does imply $K \triangleleft G$. This is a powerful result, as it provides a way to establish normality transitively under a stronger condition.
Summary of Normality Transitivity
| Property | Definition | Transitive? ($K \text{ rel } H$ and $H \text{ rel } G \implies K \text{ rel } G$) | Example/Context |
|---|---|---|---|
| Normal Subgroup ($N \triangleleft G$) | $gNg^{-1} = N$ for all $g \in G$ | No | $K={e,s} \triangleleft H={e,r^2,s,sr^2} \triangleleft D_4$, but $K \not\triangleleft D_4$ |
| Characteristic Subgroup ($K \text{ char } G$) | $\phi(K) = K$ for all automorphisms $\phi$ of $G$ | Yes (specifically, if $K \text{ char } H$ and $H \triangleleft G$, then $K \triangleleft G$) | The center of a group ($Z(G)$) or the commutator subgroup ($[G,G]$) are characteristic subgroups. |
Practical Insights and Implications
- Avoid Assumptions: Always be cautious not to assume transitivity when dealing with normal subgroups. It is a common mistake for those new to abstract algebra.
- Hierarchical Structure: Group theory often involves complex hierarchical structures of subgroups. Understanding when properties like normality propagate up the hierarchy is crucial.
- Advanced Group Theory: The concepts of normal and characteristic subgroups, and their transitivity properties, are fundamental to classifying groups and studying their internal structure, especially in areas like solvable and nilpotent groups.
