Proving a group with five elements is abelian (commutative)

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.

Answer 4

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??