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Prove that a group with $3$ elements is cyclic.

I tried the case where $G={e,a,b}
$
and I kept trying multiplication
and finally I found that
$a^2$ must equal to $b$ and $b^2$ must equal to $a$.
Then $a^3=e$.

Are there any other methods
?
I have another question :

Prove that a group with $4$ elements may or may not be cyclic.

Instead of “keeping multiplying” it's easier to fill the Cayley diagram: in every row and column every element must appear.

begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & \
b & b & \
end{array}
In the slot corresponding to $a^2$ you cannot put $e$, because otherwise the slot for $ab$ would contain $b$. Hence $a^2=b$
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b\
b & b & \
end{array}
and now the diagram has a unique completion
begin{array}{c|ccc}
& e & a & b \ hline
e & e & a & b \
a & a & b & e \
b & b & e & a\
end{array}
Then $a^3=a^2a=ba=e$.

You can try your hand with a four element group and see that the diagram admits different completions.

begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & \
b & b & \
c & c &
end{array}
In the slot for $a^2$ we can put any of $e$, $b$ or $c$. Let's try with $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e\
b & b & \
c & c &
end{array}
Then we are forced to put in the following
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c\
c & c & b
end{array}
For $b^2$ we can have either $e$ or $c$. Let's try $e$:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e\
c & c & b &
end{array}
Now we can complete:
begin{array}{c|cccc}
& e & a & b & c \ hline
e & e & a & b & c \
a & a & e & c & b \
b & b & c & e & a\
c & c & b & a & e
end{array}
It would be very tedious to verify that this diagram indeed produces a group. It does and is the product of two cyclic groups of order two; it is also the subgroup of $S_4$ given by ${id,(12)(34),(13)(24),(14)(23)}$.

According to Lagrange's theorem, a group with 3 elements can have only the trivial subgroups. Because 3 is prime only 1 and 3 divide it. So, only possible subgroups are ${e}$ with one element and $G$ with three. For $x in G$ that is not $e$ a group generated with this element,
$<x>$ must be whole $G$. So $G$ is cyclic.

A group with 4 elements can have untrivial subgroups as 4 isn't prime. For instance $mathbb{Z_4} = {0, 1, 2, 3}$ with $+$ is cyclic. It is generated with the element $1$. But group $mathbb{Z_2} times mathbb{Z_2} = {(0, 0), (1, 0), (0, 1), (1, 1)}$ with $+$ defined as $(a, b) + (a', b') = (a + a', b + b')$ is not cyclic.