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If $B \in S_7$ and $|B^3| = 7$, prove that $|B|=7$.

Solution: As $o(B^k) = o(B) / (o(B),k) $ Thus $|B| / (|B|,3) = 7$ Let $|B| = 7a$. Then $7a/(7a,3)$ should be $7a/a = 7$ or $(7a,3) = a$.

As $3$ is a prime number and can only be factored into $3$ and $1$. So $a = 3$ or $a=1$. So $a = 3$ or $a = 2$.

So $|B|$ $=$ $7$ or $21$.

My question is how to remove the possibility of $B = 21$.

George J. Adams
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1 Answers1

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You have to use the known facts special to symmetric groups. Any element is a product of disjoint cycles; and the least common multiple of the lengths of these cycles is the order of that element. Now work backwards and see what it would mean for $B\in S_7$ to be of order 21.

P Vanchinathan
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