Page 49 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes For example, if G is abelian, then Z(G) = G.
We will now show that Z(G)  G.

Theorem 2: The centre of any group G is a subgroup of G.
Proof: Since e  Z(G), Z(G)  . Now,
a  Z(G) ax =xa  x  G.

x = a xa  x  G, pre-multiplying by a .
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 xa = a x  x  G, post-multiplying by a .
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 a  Z(G).
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Also, for any a, b  Z(G) and for any x  G, (ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x (ab).
 ab  Z(G).
Thus, Z(G) is subgroup of G.

3.2 Properties of Subgroups

After discussing the term subgroup let us start understanding the important properties of
subgroup.
Theorem 3: Let G be a group, H be a subgroup of G and K be a subgroup of H. Then K is a
subgroup of G.

Proof: Since K  H, K   and ab  K  a, b  K. Therefore, K  G.
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Let us discuss at subgroups of Z, in the context of Theorem 3.

Example: In earlier example we have seen that my subgroup of Z is of the form mZ for
some m  N. Let mZ and kZ be two subgroups of Z. Show that rnZ is a subgroup of kZ iff k | m.
Solution: We need to show that mZ kZ k | m. Now mZ  kZ  m  mZ  kZ  m  kZ
 m = kr for some r  Z  k | m.
Conversely, suppose k | m.
Then, m = kr for some r  Z. Now consider any n  mZ, and let t  Z such that n = mt.

Then n = mt = (kr) t = k(rt)  kZ.
Hence, mZ  kZ.
Thus, mZ  kZ iff k | m.

Theorem 4: If H and K are two subgroups of a group G, then H  K is also a subgroup of G.
Proof: Since e  H and e  K, where e is the identity of G, e  H  K.

Thus, H  K  .
Now, let a, b  H  K. By Theorem 1 , it is enough to show that ab  H  K. Now, since a, b
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 H, ab  H. Similarly, since a, b  K, ab  K. Thus, ab  H  K.
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Hence, H  K is a subgroup of G.

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