Suppose G is a cyclic group and , G are such that order() = 12 and order () = 21. Then the order of the smallest group containing and is ______________Correct answer is '84'. Can you explain this answer?

Question

Suppose G is a cyclic group and ,   G are such that order() = 12 and order () = 21. Then the order of the smallest group containing  and  is ______________Correct answer is '84'. Can you explain this answer?

Answer

Explanation:

Given, G is a cyclic group and order(G1) = 12, order(G2) = 21

Let G1 = a and G2 = b, where a and b are generators of G1 and G2 respectively.

Now, we need to find the smallest group containing G1 and G2.

As G1 and G2 are cyclic groups, they are abelian.

So, the smallest group containing G1 and G2 must be abelian.

Let G be the group containing G1 and G2.

As G1 and G2 are cyclic, it follows that G is also cyclic.

Let G = c.

Then, order(c) = LCM(order(a), order(b)) = LCM(12, 21) = 84.

Therefore, the order of the smallest group containing G1 and G2 is 84.

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