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 $1$.

Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]

Sylow Subgroups of a Group of Order 33 is Normal Subgroups
Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.
Hint.
We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]

Group of Order 18 is Solvable
Let $G$ be a finite group of order $18$.
Show that the group $G$ is solvable.
Definition
Recall that a group $G$ is said to be solvable if $G$ has a subnormal series
\[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\]
such […]

Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4
Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.
Hint.
Use Sylow's theorem.
(See Sylow’s Theorem (Summary) for a review of Sylow's theorem.)
Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is […]

If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup
Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]

Non-Abelian Group of Order $pq$ and its Sylow Subgroups
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]

A Simple Abelian Group if and only if the Order is a Prime Number
Let $G$ be a group. (Do not assume that $G$ is a finite group.)
Prove that $G$ is a simple abelian group if and only if the order of $G$ is a prime number.
Definition.
A group $G$ is called simple if $G$ is a nontrivial group and the only normal subgroups of $G$ is […]

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Let $G$ be a finite group. Let $S$ be the set of elements $g$ such that $g^5=e$, where $e$ is the identity element in the group $G$.
Prove that the number of elements in $S$ is odd.
Proof.
Let $g\neq e$ be an element in the group $G$ such that $g^5=e$.
As […]

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

Dear Saikat Bisai,

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

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

Dear anish.ray,

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