A group with five elements is Abelian

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 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 depends on what facts you have available, but you can see this quickly if know the easily deduced fact that in a group $G$ of order $n$ the order of every element is a factor of $n$. So, every element of a group of order $5$ has order $1$ or $5$, and only the identity has order $1$, by definition. Then, any other element $g$ has order $5$, so every element is a power of $g$, and so the group is abelian, and in fact $G \cong \mathbb{Z}_5$. NB this argument only used that $5$ is prime, so it works as well for any prime order.