NORMAL SUBGROUPS (OR INVARIANT SUBGROUPS OR SELF CONJUGATE SUBGROUPS)

A subgroup H of a group G is called a normal subgroup of G if every left coset of H in G is equal to the corresponding right coset of H in G.
i.e.,           aH = Ha,       ∀ a ∈ G.

If the composition defined on G be addition, then H will be a normal subgroup of G iff.
                a + H = H + a,        ∀ a ∈ G.

In general, if H is a subgroup of a group G, then the left coset aH of H in G may not be equal to the corresponding right coset Ha. In this section, our aim is to study a particular class of subgroups H for which each left coset of H in G is equal to the corresponding right coset of H in G. We call such subgroups as normal subgroups.

Remark: (i) When G is an abelian group. Then every subgroup H of G is a normal subgroup, for 
                           aH = Ha,        ∀ a ∈ G.
(ii) The subgroups {e} and G of any group G are always normal subgroups of G. These are called trivial normal subgroups.
(iii) If H is a normal subgroup of G, then we write it as 
H Δ G.

Properties:

• Normality is preserved upon surjective homomorphism, and is also preserved upon taking inverse images.
• Normality is preserved on taking direct products.
• Every subgroup of index 2 is normal.
• A normal subgroup of a central factor is normal. In particular, a normal subgroup of direct factor is normal.

                                  Examples

Example: Let G = S₃, the symmetric group on three numbers 1, 2, 3.Show that the subgroup 
                           H = {i, (123), (132)}
is a normal subgroup of G but the subgroup
                           K = {i, (12)}
is not a normal subgroup of G.
Sol. We know that Ha = H = aH   if a ∈ H.
Since i, (123), (132) ∈ H
∴ iH = Hi, (123)H = H(123), (132)H = H(132).
Now       (12)H = {(12)i, (12)(123),(12)(132)}
                          = {(12)(23)(13)} .
and        H(12) = {i(12), (123)(12), (132)(12)}
                          = {(12)(13)(23)}.
∴           (12)H = H(12).
Again    (23)H = {(23)i, (23)(123),(23)(132)}
                          = {(23)(13)(12)} .
and        H(23) = {i(23), (123)(23), (132)(23)}
                          = {(23)(12)(13)}.
∴           (23)H = H(23).
Also       (13)H = {(13)i, (13)(123),(13)(132)}
                          = {(13)(23)(12)} .
and        H(13) = {i(13), (123)(13), (132)(13)}
                          = {(13)(12)(23)}.
∴           (13)H = H(13).
Thus      xH = Hx,       ∀ x ∈ S₃.
∴  H is a normal subgroup of S₃.
But        (13)K = {(13)i, (13)(12)}
                         = {(13)(132)}
and        K(13) = {i(13), (12)(13)}
                         = {(13)(123)}
Clearly   (13)K ≠ K(13).
Hence K is not a normal subgroup of G.

Example: If H is a subgroup of G of index 2 in G. Then H is normal subgroup of G.
Sol. Let H be a subgroup of G such that [G : H] = 2.
∴  The number of distinct left (or right) cosets of H in G is 2.
To show that H is a normal subgroup of G.
It is sufficient to prove that xH = Hx,    ∀ x ∈ G.
Let x ∈ G be arbitrary element of G.

Case I. When  x ∈ H.
Since x ∈ H           So,    xH = H = Hx
Hence  xH = Hx.

Case II. When x ∉ H.
∴          xH ≠ H         and   Hx ≠ H.
Also     [G : H] = 2.
∴         H ∪ xH = G = H ∪ Hx
⇒         xH = Hx.
Combining the two cases, we find that
            xH = Hx      ∀ x ∈ G.
∴   H is a normal subgroup of G.