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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.

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Hint: If the group has an element of order $55$ you are done, so you can assume it has one of order $5$ and it must then have how many element of order $5$? The subgroups of order $5$ intersect trivially, so what are possible numbers of elements of order $5$? – Tobias Kildetoft Apr 19 '13 at 15:06
up vote 7 down vote accepted

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.)

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How do you get that there would be 34 elements remaining with order 5? – AlanH Jun 9 '13 at 4:28
There are only four possible orders of elements: 1, 5, 11 and 55. We already eliminated 55. If 20 are order 11 by assumption, and one is order 1 (always), that leaves $55-20-1=34$ elements of order 5. – rschwieb Jun 9 '13 at 12:03

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.

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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)

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