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 $0\lt a_1 \le a_2 \le ...$. We see $z$ must be a multiple of $p$, but since $z\ge 1$ 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 $H \triangleleft G$ 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 $s \in G \setminus H$. 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 $D_{2n}: \{ a, c | a^n = c^2 = (a c)^2 = 1 \}$ the cyclic subgroup $\langle a \rangle$ 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

\[ \array { K_0 &=& \{1\} \\ K_1 &=& \{(1 2),(1 3),(1 4), (2 3), (2 4), (3 4)\} \\ K_2 &=& \{(1 2 3),(1 2 4),(1 3 2), (1 3 4), (1 4 2), (1 4 3), (2 3 4), (2 4 3)\} \\ K_3 &=& \{(1 2)(3 4),(1 3)(2 4),(1 4)(2 3) \} \\ K_4 &=& \{(1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2)\} } \]

It can be verified that $V = K_0 \cup K_3$ forms a subgroup thus is normal.