If S7 denote the symmetric group of all permutation of { 1 2 3 4 5 6 7}

pick the true statement

 

a) S7 has an element of order 10.

b) S7 has an element of order 15.

c) the order of any element of S7 is atmost 12.

how to do it?

If S7 denote the symmetric group of all permutation of { 1 2 3 4 5 6 7}

pick the true statement

 

a) S7 has an element of order 10.

b) S7 has an element of order 15.

c) the order of any element of S7 is atmost 12.

how to do it?

 

 

Apply the theorems that you should know that pertain to finite groups.

You have to use properties of [math]S_7[/math]. In particular, every [math]s \in S_7[/math] can be written as the product of disjoint cycles, and then the order of s is the lcm of the cycle lengths.

Example: [math]s=(1 2 3)(5 6)[/math]. Since the cycles are disjoint, we have [math]o(s)=lcm(3,2)=6[/math].

 

Now consider any element [math]s \in S_7[/math]. If we wanted the order of s to be 20 (for example), we'd only be allowed to use cycle lenghts 1, 2, 4, 5, 10, 20. Since 10 and 20 are too large (we're in [math]S_7[/math] after all), that means we can only use cycle lengths 2, 4 and 5 (not counting 1 as a cycle length). But out of those, we'd have to use both 4 and 5 (to obtain a lcm of 20) - but we can't, since the cycles have to be disjoint and [math]4+5>7[/math]. So there is no element of order 20 in [math]S_7[/math]. You do the rest.