Show that a group of order $5$ is always abelian. 
Show that a group of order $5$ is always abelian.

I know that if the binary operation $*$ is defined on a group $G$ is commutative (i.e. $a*b=b*a\ \forall a,b\in G$),\ then G is called a commutative(or abelian) group.
But I don't know how to solve a group of order $5$ is abelian.
Thanks in advance!  
 A: Hint: Take some non-identity element $a\in G$. Show that the set $\{a^n : n\in\mathbb N\}$ is a subgroup of $G$, and that it has $5$ elements so is the whole group $G$. Thus every element of $G$ is a power of $a$, so it is easy to show that $G$ is abelian.
A: You say in comments that "you know" that every group of prime order is cyclic, and that every cyclic group is abelian. I do not understand how this can give you "only" the validity, but not a proof. What's "only" about it?
If you already know that a group $G$ is cyclic, then you can prove explicitly that it is abelian as follows: $G$ is cyclic means that there exists $g\in G$ such that for every $x\in G$ there exists (at least one) $n\in\mathbb{Z}$ such that $x=g^n$.
So if $x,y\in G$, then there exist $n,m\in\mathbb{Z}$ such that $x=g^n$ and $y=g^m$. Then $xy = g^ng^m = g^{n+m} = g^{m+n} = g^mg^n = yx$, which proves that $xy=yx$ for all $x,y\in G$, hence $G$ is abelian.
If you already know that every group of prime order is cyclic, then a group of order $5$ is cyclic, and by the argument above must be abelian.
If you "know" that every group of prime order is cyclic but you don't know how to prove it (in which case I would not describe it as something you "know", but that's just me), then it becomes a little trickier.  I will assume you don't know Lagrange's Theorem:

Lagrange's Theorem. If $G$ is a finite group of order $n$, and $H$ is a subgroup of order $d$, then $d$ divides $n$.

A corollary is: if $G$ is a finite group of order $n$, and $g\in G$, then the order of $g$ is a divisor of $n$.
Using Lagrange's Theorem, we have: let $G$ be a group of order $5$. Let $x\in G$, $x\neq 1$ (the identity of the group). Then the order of $x$ must divide $|G|=5$, hence $|x|=1$ or $|x|=5$. If $|x|=5$, then the cyclic subgroup generated by $x$ equals all of $G$, so $G$ is cyclic (and hence abelian as above). If $|x|=1$, then $x=1$, which we explicitly discounted. So $G=\langle x\rangle$, and we are done.
We can prove Lagrange's Theorem as follows: let $G$ be a finite group, and let $H$ be a subgroup. For every $y\in G$, let $Hy$ be the subset of $G$ given by:
$$Hy = \{hy\mid h\in H\}.$$
Claim: Let $y$ and $z$ be elements of $G$. If $Hy\cap Hz\neq\varnothing$, then $Hy=Hz$.
Proof. Let $g\in Hy\cap Hz$. Then there exist $h_1,h_2\in H$ such that $g=h_1y=h_2z$. In particular, $zy^{-1}=h_2^{-1}h_1\in H$, and $yz^{-1}=h_1^{-1}h_2\in H$.  We prove that $Hy=Hz$ by double inclusion.
Let $g\in Hz$. Then $g=hz$ for some $h\in H$. Note that $hh_1^{-1}h_2\in H$, so $(hh_2^{-1}h_1)y\in Hy$. But 
$$(hh_2^{-1}h_1)y = h(h_2^{-1}h_1)y = h(zy^{-1})y = hz(y^{-1}y) = hz= g,$$
so $g\in Hy$. Thus, $Hz\subseteq Hy$. Symmetrically, if $g=hy\in Hy$, then 
$$g = hy = hyz^{-1}z = h(h_1^{-1}h_2)z = (hh_1^{-1}h_2)z\in Hz,$$
so $Hy\subseteq Hz$. Thus, if $Hy\cap Hz\neq\varnothing$, then $Hy=Hz$.
Claim 2. For all $y\in G$, $y\in Hy$.
Proof. We can write $y=1y$, with $y\in H$.
Claim 3. The sets $Hy$, $y\in G$, are a partition of $G$.
Proof. Their union is everything, since $y\in Hy$ for all $y\in G$; and any two distinct sets are disjoint by Claim 1. So they form a partition.
Claim 4. If $y\in G$, then $|Hy|=|H|$.
Proof. Define $f\colon H\to Hy$ by $f(h) = hy$. This is one-to-one, since $hy=h'y$ implies, by cancellation, $h=h'$; and onto, since every element of $Hy$ is of the form $hy$ for some $h\in H$. Thus, $f$ is a bijection.
Proof of Lagrange's Theorem. Let $|G|=n$ and $|H|=d$. Let $g_1,\ldots,g_k$ be elements of $G$ such that $Hg_i\cap H{g_j}=\varnothing$ if $i\neq j$, and with $G=\cup_{i=1}^k Hg_i$. They exist since the $Hg$ form a partition of $G$ and $G$ is finite. Then
$$n=|G| = \left|\cup_{i=1}^k Hg_i\right| = \sum_{i=1}^k|Hg_i| = \sum_{i=1}^k|H| = k|H| = kd,$$
so $d$ divides $n$ as desired. $\Box$
And now you have all the ingredients.
