For Contemporary Abstract Algebra 4 is Let G be a group Let

For Contemporary Abstract Algebra

**(4) is Let (G, ?) be a group. Let a, b ? G such that a ? b = b ? a,|a| = n, and |b| = m, where n, m ? N. Prove that |a ? b| = lcm(n, m).

(5) Let (G,*) be a group and a,a2an E G such that a, commutes with a, for all *) be a group and a1, a2, . . . , In E G su 12-n i and j. Assume also that ai has order mi for i. Use mathematical induction and i and j. Assume also that ai has order mi for all i. Use mathematical induction and (4) to prove that lai * a2 * . . . * anl = 1cm(mi, m2, , mn).

Solution

G is a commutative group with ai*aj = aj*ai for all i and j from 1 to n.

a1 has order m1 and a2 has order m2,...an has order mn.

Consider only two elements a1 and a2

Let a1*a2 = a\'

a1 has order m1 and a2 has order m2.

If m\' is the lcm of m1 and m2

then (a\')^m\'= (a1*a2)^m\'

=( a1*a2)*( a1*a2)*( a1*a2)*...m\'times

Since commutative we can write this as

= a1^m\'*a2^m\'

Since m\' is a multiple of m1 and m2 (least common)

we get

a1^m\'*a2^m\' = e^k * e^k1 = e

Hence for two elements order of a1*a2 = lcm of m1, m2

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Assume true for n =k i.e.

Order of a1*a2*...*ak (say ak\')= lcm of m1,m2...mk (say mk\')

Now to prove for n =k+1

Let us consider ak\' and ak+1 with orders m_k\' by previous assumption and m_k+1

Since for two elements it is true we have

Order of a1*a2*...ak+1 = lcm of m_k\' and m_k+1

₌ LCM of m1, m2...m_k+1

So if true for n =k, true for n =k+1

Already proved that true for n =2

Hence by principle of mathematical induction, true for all integers upto m.

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