Can a group of order $55$ have exactly $20$ elements of order $11$? Give a reason for your answer
by sylow theorem the answer is easy but without using sylow how can I solve this.can anyone help me please.thanks for your kind help.
First, discount that could be a cyclic group, for then there would be exactly 10 elements order 11.
If the group did have 20 elements order 11, then there would be 34 remaining elements with order 5.
Counting distinct prime powered elements is easy, since the subgroups only intersect at the identity: just note that there are $p-1$ elements of order $p$ in each distinct subgroup of order $p$, and so if you know there are $n$ such subgroups, there are $n(p-1)$ elements order $p$.
But the number of order 5 elements would have to be a multiple of $4$ (which it is not, if it is 34.)
A theorem of Frobenius says the number of elements in a finite group that satisfy $x^n=e$ is a multiple of n. For n=11 this means the number of elements of order 11 plus the identity element is a multiple of 11, ie, the number of elements of order 11 is one less than a multiple of 11. So not 20.
I know you don't required Sylow method
But if we do this question by Sylow method
55= 5 x 11
there is only one Sylow 11 subgroup
so number of element of order 11 = 10 *1 =10
In case of any error (plz suggest)