A group with five elements is Abelian

I tried to prove the following theorem: A group with five elements is abelian.

I know only the definition of a group and a subgroup but no more.(this is a problem from Topics in Algebra by IN Herstein). I tried a few things but to no avail.

Suppose that the group is $G=\{e,g_1,g_2,g_3,g_4\}$. $g_1g_2$ cannot be $g_1$ or $g_2$ or else one of them will be the identity,$e$.So, $g_1g_2$ is either $g_3$ or $g_4$ or $e$.

Case I: $g_1g_2=e$.(so, $g_2g_1=e$). Then $g_1g_3=g_2$ or $g_4$($g_1g_3\not=e$,or else $g_2=g_3$). Say,$g_1g_3=g_2$. That is, $g_1=g_3^{-1}g_2\implies g_1g_2=g_3^{-1}g_2^2=e$ or in other words, $g_2^2=g_3$. Using this, $g_3g_1=g_2^2g_1=g_2(g_2g_1)=g_2$ and we have $g_1g_3=g_1g_2^2=g_2$.

Maybe, this will lead to a solution but I was wondering if there is a much smarter way to do it as this leads to a lot of casework! .

Hint: suppose your group is not abelian. Then you can find two different elements, say (after renumbering) $g_1$ and $g_2$, not equal to the identity, such that $g_1g_2 \neq g_2g_1$. Then the elements $\{e, g_1,g_2, g_1g_2, g_2g_1\}$ are all different. Now try to derive a contradiction (look at $g_1^2$ - which element of the set is this? Do the same for $g_1g_2g_1$).

Below are more details.

In a finite group, there's no non-trivial element such that $g_1^2=g_1$. Also $g_1^2$ cannot be neither $g_1g_2$ nor $g_2g_1$, because $g_2\neq g_1$. Since $g_2=g_1^2$ would yield $g_1g_2=g_1^3=g_2g_1$, it's imposible, too. Therefore $g_1^2=e$. We now observe $g_1g_2g_1$. Note that $g_1,\, g_1g_2,\, g_2g_1\neq e$, with operating on both left and right, $g_1g_2g_1\neq g_1g_2,\, g_2g_1,\, g_1$. However, if $g_1g_2g_1=e$, we'll have $g_1g_2=g_1^{-1}=g_2g_1$; if $g_1g_2g_1=g_1$, $g_1g_2=e=g_2g_1$ will be produced. A contradiction is hence derived.

Proposition 1: Any group of prime order is cyclic.

Let $G$ be a non trivial group of order $p$ and take $g\in G$, $g\neq1$. So $\langle g\rangle$ is a subgroup of $G$, hence its order must divide $p$, which is prime, so $|\langle g\rangle|=p$, hence $\langle g\rangle=G$.

Proposition 2: Any cyclic group is abelian.

A finite cyclic group is by definition of the type $G=\langle g\rangle:=\{1,g,g^2,\dots,g^{n-1}\}$. The fact its elements commute follows easily.

It is known that the order of a conjugacy class from a group element divides the order of the group. With $|G|=5$, this results either in one conjugacy class with $5$ elements, or five conjugacy classes with one element. Since the identity element always has its own conjugacy class, the first case is impossbible. Because all conjugacy classes have only one element, is $G$ commutative.