This is an exam question I found. and was the last part on a question about finding the orders of permutation cycles. So I'm thinking this is to do with the fundamental theorem of finitely generated abelian groups so element $|\sigma_1|$=16 has factorisation $2^4$ 70 has the factorisation $2\times5\times7$ as these are cyclic elements they are indeed abelian. But i don't know why $2^4$ would imply ther is an element $2\times5\times7=70$.
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2This is likely to be a duplicate question though . . . – Shaun May 12 '24 at 12:01
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$70=2\times5\times7$ – J. W. Tanner May 12 '24 at 12:02
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yeah I got the $70=2\times 5 \times 7$ thanks , but why would that imply an element of 16 would relate to this? – Rich C May 12 '24 at 12:15
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2How large is $n$ if $S_n$ has an element of order $16$, and what could an element of order $70$ look like? – J. W. Tanner May 12 '24 at 12:28
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After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. – Shaun May 12 '24 at 13:14
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2Not really a duplicate, but I was not able to find better: https://math.stackexchange.com/questions/493259 and https://math.stackexchange.com/questions/1576417 – Anne Bauval May 12 '24 at 14:09
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The divisors of $16$ are powers of two, so each of their $\operatorname{lcm}$s would be less than $16$ (e.g., $\operatorname{lcm}(2,8)=8$) unless you have just $16$. So the element $\sigma$ of order $16$ contains at least one cycle of length $16$. Thus $n\ge 16$. But $70=2\times 5\times 7$ and $2+5+7=14<16$, so all you need is an element of cycle type $(2,5,7)$, which has order $70$.
Shaun
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2The element of order $16$ could have more than one cycle of length $16$ (if $n\ge32$) – J. W. Tanner May 12 '24 at 12:59
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1But the main thing is there is no cycle lower than 16 and you can carry on with shaun's proof. – Rich C May 12 '24 at 13:09
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So is it true an element of order 2^k can have no cycle lower than 2^k which is a handy thing to know. – Rich C May 12 '24 at 13:56
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2@RichC This is false. For instance $(1234)(56)$ is of order $4$ and has a $2$-cycle. Shaun's answer only uses the fact that an element of order $2^k$ has at least one $2^k$-cycle. – Anne Bauval May 12 '24 at 14:04
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