Page 68 - DMTH403_ABSTRACT_ALGEBRA

Unit 5: Normal Subgroups

Let us consider an example. Notes

Example: Show that every ‘subgroup of Z is normal in Z.
Solution: As you know that if H is a subgroup of Z, then H = mZ, for some m  Z. Now, for any
z  Z,
H + z = { ..., - 3 m + z , - 2 m + z , - m + z , z , m + z , 2 m + z ,...}
= { ..., z - 3m, z – 2m, z – m, z, z + m, z + 2m,} {since + is commutative)
= z + H.

 H  Z.
Above example is a special case of the fact that every subgroup of a commutative group is a
normal subgroup.
Let us now prove a result that gives equivalent conditions for a subgroup to be normal.
Theorem 1: Let H be a subgroup of a group G. The following statements are equivalent.

(a) H is normal in G.
-l
(b) g Hg  H   G.
g
g
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(c) g Hg  H   G.
Proof: We will show that (a)  (b)  (c)  (a). This will show that the three statements are
equivalent.

(a)  (b) : Since (a) is true, Hg = gH  g E G. We want to prove (b). For this, consider
g Hg for g E G. Let g hg  g Hg.
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-1
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Since hg  Hg = gH, 3 h  H such that hg = gh .
1
1
 g hg = g gh = h  H
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1
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 (b) holds.
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Note g Hg = {g hg | h  H }

(b)  (c) : Now, we know that (b) holds, i.e., for g  G, g Hg  H. We want to show
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that H  g Hg. Let h  H. Then
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h = ehe = (g g) h (g g)
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= g (ghg ) g
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= g { (g ) hg } g  g Hg, since (g ) hg  (g ) H(g )  H.
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-1 -1
-1 -1
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-1 -1
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 H  g Hg.
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 g Hg = H  g  G.
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