If an abelian group has subgroups of orders m and n respec tively, then show that it has a subgroup whose order is the Icm. of m and n.
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- Lemma: Let G be an abelian group and H, K its subgroups such that o(H)=m, o(k|) =n, and H Π K={1}. Then the o(H V K)=mn where H V K is the smallest subgroup containing both H and K
- Proof : Because G is abelian,
- H V K={hk | h€H and k€K}. Let a, b€ H and x, y€ k such that ax= by. Then a-¹b=xy-¹€ H Π K and thus a-¹b=1
- Consequently, a=b and x=y.
- One possibility is to use the formula, o(HK) =o(H) • o(k) / o(H Π K),
- Then, note that since H Π K ≤ H, K by Lagrange. o(H Π K) | o(H), o(k), so that o(H Π K) | gcd (o(H), o(K), and finally recall that
- (gcd) (x, y). lcm. (x, y) =xy. With this you obtain that G has a subgroup of order a multiple of the LCM, so you will have to know that the converse of Lagrange's theorem holds in a finite abelian group.
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