Page 45 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes So, by definition, (Z,+) is a subgroup of (Q,+), (R,+) and (C,+).
Now, if (H,*) is a subgroup of (G,*), can the identity element in (H,*) be different from the
identify element in (G,*)? Let us see. If h is the identity of (H,*), then, for any a  H, h * a = a * h
= a. However, a  H  G. Thus, a * e = e * a = a, where e is the identity in G. Therefore, h * a =
e * a.
By right cancellation in (G,*)w,e get h = e.
Thus, whenever (H, *) is a subgroup of (G,*), e  H.

Remark 1: (H,*) is a subgroup of (G, *) if and only if
(i) e  H,
(ii) a, b H  a * b  H,
(iii) a  H  a  H.
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We would also like to make an important remark about notation here.
Remark 2: If (H,*) is a subgroup of (G,*), we shall just say that H is a subgroup of G, provided that
there is no confusion about the binary operations. We will also denote this fact by H  G.

Now let us first discuss an important necessary and sufficient condition for a subset to be a
subgroup.
Theorem 1: Let H be a non-empty subset of a group G. Then H is a subgroup of G iff a, b  H
 ab  H.
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Proof: Firstly, let us assume that H  G. Then, by Remark l, a, b  H  a, b  H
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 ab  H.
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Conversely, since H  ,  a  H. But then, aa = e  H.
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Again, for any a  H, ea = a  H.
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Finally, if a, b  H, then a, b  H. Thus, a (b ) = ab  H, i.e.,
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H is closed under the binary operation of the group.
Therefore, by Remark 1, H is a subgroup.

Note A subgroup of an abelian group is abelian.

Example: Consider the group (C*.,). Show that
S = { z C | |z| = 1 } is a subgroup of C*.
Solution: S  , since 1  S. Also, for any z , z  S,
1
2
1
|z z | = |z | |z | = |z | |z |  1.
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–1
1
2
2
1
1
2
Hence, z z  S. Therefore, by Theorem 1, S  C*.
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l
2
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