Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We defined a group as a set $G$ with an operation $\circ$ with

1) $\forall x,y,z, \in G: x \circ ( y \circ z) = (x \circ y) \circ z$

2) $\exists e\in G: \forall x\in G : x \circ e = e \circ x = x$

3) $\forall x \in G\;\;\;\exists x^{-1} \in G: x \circ x^{-1} = x^{-1} \circ x = e$

Now I'm wondering what group fullfilling these axioms isn't abelian, because in 2) and 3) there's already some kind of commutativity.

share|cite|improve this question
Invertible $2\times 2$ matrices under matrix multiplication. – André Nicolas Jan 21 '12 at 16:22
Bijections from $\{1,2,3\}$ to $\{1,2,3\}$ under function composition. – Mikko Korhonen Jan 21 '12 at 16:23
Additionally, note that $g^ng^m=g^mg^n$ for any element $g$ and integers $m,n$. This follows from the associative and inverse laws. – dls Jan 21 '12 at 18:07
Your axiom 3) doesn't look right! – Derek Holt Jan 22 '12 at 6:50
@DerekHolt: What's wrong about axiom 3? – meinzlein Jan 22 '12 at 7:06
up vote 7 down vote accepted

The two main examples are Matrix groups, and permutation groups. I assume you know matrices and that their multiplication is non-commutative (although multiplication by a scalar matrix is commutative, and also multiplication by inverse - which explains 2 and 3).

A permutation of a set $X$ is a function $\sigma: X\to X$ which is one-to-one and unto; multiplication is defined by function composition. Try finding two such functions that do not commute for the set $X=\{1,2,3\}$.

share|cite|improve this answer

Too long for a comment:

I essentially would like to point to your wondering about some sort of "commutativity".

The two statements basically say:

  1. Every element commutes with identity element in the group.
  2. Every element commutes with its own inverse.

Looking at the kind of question, you have asked us, I assume, you have just started understanding these abstract friends called groups!

As you proceed, you'll see that all elements in the cyclic group generated by an element commute with the element. Further, all elements that commute with a particular element form a subgroup of the group, and is called the centralizer of that element. More generally, all those elements that commute with every element in $G$, form what is known as the center of the group and in some sense the largest abelian part sitting inside $G$.

As for non-abelian groups, I would suggest that you know the following groups:

  • Dihedral Groups

    The group of rigid motions of a regular polygon under composition

  • Permutation Groups

    The group of bijections from a set onto itself under composition

Have fun with Group Theory!

share|cite|improve this answer

One particularly concrete example of a non-abelian group is the Rubik's cube group..

What is this group, first of all?

Here, elements of the group are the possible sequences of moves, and multiplication of move sequences A, B is just performing sequence A, and then performing sequence B, giving a new sequence of moves. The identity element is simply the sequence with no moves (which certainly commutes), and the inverse of an element C of this group is the sequence that takes the cube back to its solved state after performing C on a solved cube.

Why is this set of moves a group?

After playing around with a physical cube or an applet such as this one, it becomes relatively clear that elements commute with their inverses, and that the group operation is associative, which, together with closure (you can't combine two move sequences and return something that isn't still a sequence of moves on the cube), mean that this is indeed a group.

Why is this non-abelian?

Note, however, that there are several elements in this group that do not commute, such as single rotations of adjacent faces, as well as several non-identity moves that are not inverses of each other, but still commute.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.