## 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

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

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

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