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While doing my homework i find out that the maximum order of an element in $S_3$ is 3 (the element $(123)$) and the maximum order of an element in $S_4$ is 4 (the element $(1234)$)

Can i generalize that and say that the maximum order of an element in $S_n$ is n?

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  • $\begingroup$ Check $S_5$ and see. $\endgroup$ – vadim123 Dec 30 '13 at 16:27
  • $\begingroup$ i asked to check if there is any theorem that prove what i wrote $\endgroup$ – Aviad Chmelnik Dec 30 '13 at 16:30
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    $\begingroup$ What you wrote isn't true; you need only look at $S_5$ for a counterexample. $\endgroup$ – vadim123 Dec 30 '13 at 16:30
  • $\begingroup$ What vadim writes is accurate, @AviadChmelnik : before asking questions it is worthwhile to really try by yourself, for example by trying a decent ammount of examples. As he tells you,in $\;S_5\;$ you already have acounter example as the element $\;(12)(345)\;$ has order $\;6>5\$, and in $\;S_8\;$ you already have elements of order $\;15\;$ , say... $\endgroup$ – DonAntonio Dec 30 '13 at 16:38
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Well after checking $S_5$ i saw that what i said is not correct: element $(123)(45)$ has order of 6. thanks @vadim123

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In general, you have to look at disjoint cycle types and hence for a "maximum" partition of $n=a_1 + \dots a_k$ and take the lcm of the $a_i$'s. Example for $S_7$: $7=2+5$ gives you order $10$ but $7=3+4$ yields $12$.

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Your answer is here.

The maximum order of an element of finite symmetric group by William Miller, American mathematical monthly, page 497-506.

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