# Determine the Number of Elements of Order 3 in a Non-Cyclic Group of Order 57

## Problem 628

Let $G$ be a group of order $57$. Assume that $G$ is not a cyclic group.

Then determine the number of elements in $G$ of order $3$.

Contents

## Proof.

Observe the prime factorization $57=3\cdot 19$.
Let $n_{19}$ be the number of Sylow $19$-subgroups of $G$.

By Sylow’s theorem, we know that
$n_{19} \equiv 1 \pmod{19} \text{ and } n_{19} \mid 3.$ It follows that $n_{19}=1$.

Now, observe that if $g\in G$, then the order of $g$ is $1$, $3$, or $19$. Note that since $G$ is not a cyclic group, the order of $g$ cannot be $57$.
As there is exactly one Sylow $19$-subgroup $P$, any element that is not in $P$ must have order $3$.

Therefore, the number of elements of order $3$ is $57-19=38$.

### Remark.

Note that there are $18$ elements of order $19$ and the identity element is the only element of order $1$.

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### 4 Responses

1. Saikat Bisai says:

I think there are 18 elements of order 19, not 16.

• Yu says:

Dear Saikat Bisai,

Thank you for catching the typo. I just fixed it.

2. anish.ray says:

The number of elements of order 19 must be 18, not 16.

• Yu says:

Dear anish.ray,

Thank you for catching the typo. I just fixed it.

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