Group Theory - Normal Subgroups

Normal Subgroups

Two elements $a, b$ in a group $G$ are said to be conjugate if $t^{- 1} a t = b$ for some $t \in G$ . The elements $t$ is called a transforming element. Note conjugacy is an equivalence relation. Also note that conjugate elements have the same order. The set of all elements conjugate to $a$ is called the class of $a$ .

Theorem: The elements of $G$ that commute with a given element $a$ form a subgroup $N$ , called the normalizer of $a$ . Given a decomposition of $G$ into cosets $N g_{1}, . . ., N g_{h}$ , where $h = ∣ G ∣ / ∣ N ∣$ , the elements of the class of $a$ can be written $g_{1}^{- 1} a g_{1}, . . ., g_{h}^{- 1} a g_{h}$ .

Proof: That the normalizer is indeed a subgroup is easily verified. If we take any $n g_{i} \in N g_{i}$ where $n \in N$ then we have

(n g_{i})^{- 1} a (n g_{i}) = g_{i}^{- 1} n^{- 1} a n g_{i} = g_{i}^{- 1} a g_{i}

Also, if we have $g_{i}^{- 1} a g_{i} = g_{j}^{- 1} a g_{j}$ then $g_{i} g_{j}^{- 1}$ also commutes with $a$ , thus also belongs to $N$ , implying that $N g_{i} = N g_{j}$ .

Note an element $a$ forms a class by itself if and only if $a$ commutes with all of $G$ . Such an element is called an invariant or self-conjugate element of $G$ . In every group, the identity is invariant. In an abelian group every element is invariant.

Classes of conjugates are disjoint, for if $g^{- 1} a g = h^{- 1} b h$ then $x^{- 1} a x = (g h^{- 1} x)^{- 1} b (g h^{- 1} x)$ for any $x \in G$ , implying that every element in the class of $a$ also belongs to the class of $b$ . Thus we may decompose $G$ into disjoint classes of conjugates, and if there are $k$ classes, we have $∣ G ∣ = h_{1} + . . . + h_{k}$ where $h_{i}$ is the size of the $i$ +++th class. Note each $h_{i}$ divides $∣ G ∣$ and $h_{i} = 1$ if and only if $a_{i}$ + is self-conjugate.

Theorem: If a group $G$ has order $p^{m}$ for some prime $p$ , then the number of self-conjugate elements is a positive multiple of $p$ .

Proof: Consider the decomposition of $G$ . Using the above notation, each $h_{i}$ must be some nonnegative power of $p$ . Then suppose $z$ of the $h_{i}$ are equal to one (so $z$ is the number of self-conjugates). Then we have

p^{m} = z + p^{a_{1}} + p^{a_{2}} + . . .

where $0 < a_{1} \leq a_{2} \leq . . .$ . We see $z$ must be a multiple of $p$ , but since $z \geq 1$ because $1$ is always invariant, $z$ must be a positive multiple of $p$ .

We may generalize some of these concepts as follows: If $K$ is a subset of some group $G$ then any subset of the form $g^{- 1} K g$ is said to be conjugate with $K$ . The elements of $G$ which commute with $K$ form a group $N$ which is the normalizer of $K$ . In a similar manner to above we can show:

Theorem: The number of sets conjugate to $K$ is the index of its normalizer $N$ .

A set $H$ that commutes with every element of $G$ is called invariant or self-conjugate. In particular, if $H$ is some subgroup of $G$ , then we call $H$ a normal or invariant or self-conjugate subgroup of $G$ . In general, if $A$ is some subgroup of $G$ then groups of the form $g^{- 1} A g$ are called the conjugate subgroups of $A$ . Write $H ◃ G$ to express that $H$ is a normal subgroup of $G$ . Note that the intersection of normal subgroups is also a normal subgroup, and that subgroups generated by invariant sets are normal subgroups.

Theorem: A subgroup of index 2 is always normal.

Proof: Suppose $H$ is a subgroup of $G$ of index 2. Then there are only two cosets of $G$ relative to $H$ . Let $s \in G ∖ H$ . Then $G$ can be decomposed into the cosets $H, s H$ or $H, H s$ , implying $H$ commutes with $s$ . Since $H h = h H$ for any $h \in H$ we see that $H$ commutes with every element of $G$ and hence is normal.

Example: In the dihedral group $D_{2 n} : {a, c ∣ a^{n} = c^{2} = (a c)^{2} = 1}$ the cyclic subgroup $⟨ a ⟩$ is normal.

Example: The alternating group $A_{n}$ is normal in $S_{n}$ .

Note if $a$ is an element of a normal subgroup $H$ of a group $G$ , then the class of $a$ is contained in $H$ , so that a normal subgroup can be viewed as the union of classes of $G$ , and conversely, any union of classes of $G$ satisfying the group axioms form a normal subgroup of $G$ .

Example: The classes of $S_{4}$ are

\begin{matrix} K_{0} & = & {1} \\ K_{1} & = & {(12), (13), (14), (23), (24), (34)} \\ K_{2} & = & {(123), (124), (132), (134), (142), (143), (234), (243)} \\ K_{3} & = & {(12) (34), (13) (24), (14) (23)} \\ K_{4} & = & {(1234), (1243), (1324), (1342), (1423), (1432)} \end{matrix}

It can be verified that $V = K_{0} \cup K_{3}$ forms a subgroup thus is normal.