# 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.

*blynn@cs.stanford.edu*💡