Proving a group with five elements is abelian (commutative) I'm very early in my study of algebra, and would help to show that every group $G$ with five elements is abelian (commutative). Preferably in a more elemental way possible, I began studying algebra a little time, and I need this example to understand better.
 A: As nik pointed out: using Lagrange's theorem we can see its only subgroup is $e$ and itself. (since 5 is prime).
So therefore the group is cyclic. (since having elements of a lower order would  create a subgroup). then the elements are $e,a^1,a^2,a^3,a^4$. since $a^k\cdot a^l=a^{k+l}=a^{l+k}=a^l\cdot a^k$ the group is abelian
A: Nicky Heckster has given a useful hint here. I am expanding on it.
Let $e$ be identity element. Suppose that there exist two elements such that $ab\neq ba$. Then 5 unique elements of the group will be $\{e,a,b,ab,ba\}$. Note that $a^2 \neq ab$ and $a^2 \neq ba$. Also, $a^2 \neq b$ and $b^2 \neq a$ as either of these will make $ab=ba$. We can conclude $a^2=b^2=e$. 
Now consider the product of $a$ with all other elements
$a.\{e,a,b,ab,ba\}=\{a,a^2,ab,a^2b,aba\}=\{a,e,ab,b,aba\}$. Note that product of $a$ with each of the elements should be unique, as $ ax=ay \implies a^{-1}ax=a^{-1}ax \implies x=y$. All elements except $ba$ are already in set of products. This means that $aba=ba \implies a=e$ But this is a contradiction. Therefore a group of 5 elements has to be commutative.
A: First, Lagrange's theorem implies there is no subgroup of order 2, so one may suppose $\{1, a, \bar{a}, b, \bar{b}\}$ be this group.
Consider $ab$, there are two cases $ab=\bar{a}$ or $ab=\bar{b}$. We obtain $a^2b=1$ or $ab^2=1$, hence abelian in both cases.
A: As $|G|=5$ there would be certainly a non identity element $a\in G$ 
Before going any further It is necessary to see that :

A Group in which every element is of order $2$ is abelian  

so, we can for sure choose non identity element $a$ to be such that $a^2\neq e$
So, we have $3$ distinct elements $\{e,a,a^{-1}\}$ from $G$
For sure $a\in G$ can not have order more than $5$. 


*

*Suppose $a$ is of order $5$ then we see that $G$ is abelian.

*$a$ can not have order $4$ because a subgroup of $G$ can not be of order $|G|-1$


So, only possibility we should take care of is $|a|=3$
In that case we have $\{e,a,a^2\}\subset G$
As $|G|=5$ there would be certainly one elemnet $b\in G$ such that $b\notin \{e,a,a^2\}$ and $b$ can not be inverse of any element in $\{e,a,a^2\}$
Only possible order of  $b$ is $2$... 

what would go wrong if $b$ is of order $3$ (consider possibilities of $ab$)

So, we have $\{e,a,a^2,b\}\subset G$
Now... What could $ab$ be?


*

*$ab=e\Rightarrow ??$

*$ab=a\Rightarrow ??$

*$ab=a^2\Rightarrow ??$

*$ab=b\Rightarrow ??$


So, $ab$ is certainly another element which should complete the group as $|G|=5$ and $|\{e,a,a^2,b,ab\}|=5$
Now, $ba$ should also be in $G$
Now... What could $ba$ be?


*

*$ba=e\Rightarrow ??$

*$ba=a\Rightarrow ??$

*$ba=a^2\Rightarrow ??$

*$ba=b\Rightarrow ??$


so, only possible choice is $ab=ba$

$\textbf{ Hold On! this does not say that $G$ is abelian}$

certainly $a^2b$ should also be in $G$ Right?
what are the possibilities for $a^2b$ ?


*

*$a^2b=e\Rightarrow ??$

*$a^2b=a\Rightarrow ??$

*$a^2b=a^2\Rightarrow ??$

*$a^2b=b\Rightarrow ??$

*$a^2b=ab\Rightarrow ??$


This adds one more element to $G$ which contradicts given condition of cardinality of  $G$ being $5$
So... What does this say??
