Page 69 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes (c)  (a) : For any g  G, we know that g Hg = H.
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 g(g Hg) = gH, that a, Hg = gH.
 H  G, that is, (a) holds.

Note Ha = Hb  Hac = Hbc for any a, b, c  G.

We would like to make the following remark about Theorem 1.
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Remark: Theorem 1 says that H  G  g Hg  H  g e G. This does not mean that
g hg = h  h  H and g  G .

For example, you have shown that A  S . Therefore, by Theorem 1,
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(1 2) A (1 2) = A . But, (1 2) (1 3 2) (1 2)  (1 3 2). In fact, it is (1 2 3).
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Theorem 2: Every subgroup of a commutative group is normal.
Proof: Let G be an abelian group, and H  G. For any g  G and h  H, g hg = (g g)h = h E H.
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 g Hg  H. Thus, H  G.
Theorem 2 says that if G is abelian, then all its subgroups are normal. Unfortunately, the converse
of this is not true. That is, there are non-commutative groups whose subgroups are all normal.
We will give you an example after doing Theorem 3. Let us first look at another example of a
normal subgroup.

Example: Consider the Klein 4-group, K , given in table below. Show that both its
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subgroups < a > and < b > are normal.

× e a b ab
e e a b ab
a a e ab b
b b ab e a
ab ab b a e

Solution: Consider the table of the operation given in table. Note that a and b are of order 2.
Therefore, a = a and b = b . Also note that ba = ab.
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Now, let H = < a > = {e, a}. We will check that H  K , that is, g hg  H  g  K and h E H.
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Now, g eg = e E H  g E K .
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Further, e ae = a  H, a aa = a  H, b ab = bab = a  H and (ab) a(ab)
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= b (a aa)b = bab = a E H.
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 H  K .
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By a similar proof we can show that < b >  K .
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