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.

  • 1
    $\begingroup$ Do you know Lagrange's theorem? $\endgroup$ Feb 18, 2014 at 13:32
  • $\begingroup$ If you understand the answers to this question, then it's easy, because a cyclic group is always abelian. $\endgroup$ Feb 18, 2014 at 13:47
  • 1
    $\begingroup$ I am sure this question is asked before proving lagrange theorem... lagrange theorem is an overkill here I believe... $\endgroup$
    – user87543
    Feb 18, 2014 at 16:10
  • $\begingroup$ @PraphullaKoushik Sounds like a job for Cayley-table-Sudoku! $\endgroup$
    – user1729
    Feb 18, 2014 at 17:10
  • $\begingroup$ @user1729 : Exactly! :) $\endgroup$
    – user87543
    Feb 18, 2014 at 17:12

4 Answers 4


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


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.


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.

  • 1
    $\begingroup$ Using an overline to denote inverse is not very common (as that notation is also often used for images in quotients). Also, it is not quite clear here why you can assume that none of the non-identity elements are equal to their inverses. $\endgroup$ Feb 18, 2014 at 13:42
  • $\begingroup$ Why does $a^2b=1$ imply that $ab=ba?$ $\endgroup$
    – 5xum
    Feb 18, 2014 at 13:43
  • $\begingroup$ @5xum $a^2 b=1$ implies $b=\bar{a}^2$. $\endgroup$
    – Ma Ming
    Feb 18, 2014 at 13:54
  • $\begingroup$ Yes, it's OK now, when you explained that $\bar a \neq a$. $\endgroup$
    – 5xum
    Feb 18, 2014 at 13:56

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??

  • 1
    $\begingroup$ The only possible orders are $1$ and $5$ by Lagrange's theorem. This is unnecessarily roundabout. $\endgroup$
    – Pedro
    Feb 18, 2014 at 16:54
  • $\begingroup$ @PedroTamaroff : See my comment just below the question....That is what my point is.... This is usually asked much before introducing Lagrange theorem... Lagrange theorem is an overkill... $\endgroup$
    – user87543
    Feb 18, 2014 at 16:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .