Properties of subgroup Algebraic structures
(Monday, May 3, 2010)
Theorem 1:
The identity element of a subgroup is the same as that of the group.
Proof:
Let S be a subgroup G under *. Then by definition, S and G are both groups under the same operation * and S Ì G. Let e and e' be the identity elements of G and S respectively.
Let a e S, then a e G. We have
a * e' = a and a * e = e for the existence of identity element of S and G.
\ a * e = a * e' which gives e = e' (by
cancellation law)
Hence, the result.
Theorem 2:
The inverse of an element of a subgroup of a group is same as that of the same element belonging to the group itself.
Proof:
Let S be a subgroup of G under * and e be their identity element. Let a e S, the a e G.
Let a' and a'' be the inverses of the element a of áS, *ñ and áG, *ñ respectively. Then we have
a * a' = e, where a' e S
and
a * a'' = e, where a'' e G.
\a * a' = a * a'‘; for a, a', a'' e G.
Þ a' = a'' (by cancellation law)
Hence, the result.
Theorem 3:
If áG, *ñ be a group and SÌG, the S will be a subgroup of G if and only if
(i) a, b e S Þ a * b e S
(ii) a e S Þ a-1 e S
Proof:
If S be a subgroup of G under *, then S itself is a group under *.
Hence, it is evident by the axiom G1 and G4 that
(i) a, b e S Þ a* b e S
(ii) a e SÞ a-1 e S.
Conversely, suppose that S Ì G and conditions (i) and (ii) hold. In order to establish that S is a subgroup of G, all that is needed is to verify that e e S,
where e is the identity element of G and that the associative law hold for the element of S.
Since, the associative law hold in G and SÌG, the law hold all the more for elements of S (the property being hereditary).
Also, by (ii), a eS Þa-1e S and then by (i) e=a * a-1 e S. This completes the proof.
Theorem 4:
A subset S of a group G under the binary operation * is a subgroup of
áG, *ñ, if and only if
a e S and b e S Þ a-1 * b e S.
Proof:
If S be a subgroup of G under *, then the elements a, b e S Þ a-1 e S and hence a-1 * b e S.
Conversely, let a, b e S Þ a-1 * b e S.
Then a-1 * a = e e S which gives that
a-1 * e = a-1 e S.
\ (a-1)-1 * b = a * b e G.
Hence, S is a subgroup under the operation *.
Example:
Prove that the intersection of any two subgroups of a group is a subgroup of the group.
Proof:
Let G be any group and H1 and H2 be its subgroups under a binary operation *. The intersection H1 Ç H2 is non-empty since at least
the identity element is common to both.
Let a, b e H1ÇH2. Then a, b e H1 and H2.
Again as a, b e H1 and H1 is a subgroup of G, therefore, a-1 * b e H1. For the same reason a-1 * b e H2.
\ a-1 * b e H1ÇH2
Thus, a, b e H1 Ç H2 Þ a-1 * b e H1ÇH2 and (H1Ç H2) Ì G.
Hence, it follows that H1 Ç H2 is a subgroup of G.
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