# maximum order of an element in symmetric group [duplicate]

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?

• Check $S_5$ and see. – vadim123 Dec 30 '13 at 16:27
• i asked to check if there is any theorem that prove what i wrote – Aviad Chmelnik Dec 30 '13 at 16:30
• What you wrote isn't true; you need only look at $S_5$ for a counterexample. – vadim123 Dec 30 '13 at 16:30
• 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... – DonAntonio Dec 30 '13 at 16:38

Well after checking $S_5$ i saw that what i said is not correct: element $(123)(45)$ has order of 6. thanks @vadim123
• @Hagen: LOL! This is how physicists (no offence!) prove that $60$ is divisible by any number: it is divisible by $1,2,3,4,5,6, ...$ – Nicky Hekster Dec 30 '13 at 16:46
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 $12$.