For an arbitrary n in Z, the cyclic subgroup \langle n \rangle of Z, generated by n under addition, is the set of all multiples of n. Describe the subgroup \langle m \rangle \cap \langle n \rangle for arbitrary m and n in Z.

Armorikam

Armorikam

Answered question

2021-08-04

For an arbitrary n in Z, the cyclic subgroup n of Z, generated by n under addition, is the set of all multiples of n. Describe the subgroup mn for arbitrary m and n Z.

Answer & Explanation

bahaistag

bahaistag

Skilled2021-08-05Added 100 answers

Given information:
For an arbitrary n in Z, the cyclic subgroup n of Z generated by n under addition, and it is the set of multiples of n.
Calculation:
For an arbitrary n in Z, the cyclic subgroup m of Z generated by m under addition, and it is the set of multiples of m.
mn is the subgroup that contains multiples of least common multiple of m and n.
Consider the groups generated by mn and l where l.c.m(m,n) = l.
Let xmn
xm and xn then x=mk1 and x=nk2 for some k1,k2Z.
Thus, mx and nx as l.c.m.(m,n)=llx.
Therefore, xl
Thus, mnl.
Let yl. then y=lk.
As l is least common multiple of m and n,ml and nl.
Thus, l=ma and l=mb for some a,bZ.
y=ma and y=nb.
Hence ymn.
Thus, lmn.
Therefore,
mn=l.
Therefore, the elements in the subgroup mn for arbitrary m,nZ are generated by
l.c.m(m,n).

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