I am reading Abstract Algebra. I cannot visualise the following example:

Let $n$ be a positive integer, and consider the set $S_n$ of all permutations from the set $n = {1, 2, \ldots , n}$ to itself. Let $n = 3$, and consider the group $S_3$ . Two of the permutations in this group are $\phi_1$ and $\phi_3$ , where $\phi_1$ sends $1$ to itself and transposes $2$ and $3$, and $\phi_3$ sends $3$ to itself and transposes $1$ and $2$.

Let's apply the group operation to this pair of permutations, looking at $\phi_1\circ \phi_3$ and $\phi_3\circ \phi_1$ . The effect that $\phi_1\circ \phi_3$ has on $1$ is $(\phi_1\circ \phi_3)(1) = \phi_1(\phi_3(1)) = \phi_1(2) = 3$, but the effect that $\phi_3\circ \phi_1$ has on $1$ is $(\phi_3\circ \phi_1)( 1 ) = \phi_3(\phi_1(1)) = \phi_3(1) = 2$. Hence $\phi_1\circ \phi_3 \neq \phi_3 \circ \phi_1$.

I did not understand what happened here.

My Attempt: $S_3$ should have $6$ elements. What are $\phi_1$ and $\phi_3$ here?

I don't see how/why $\phi_3(1)=2$ and $\phi_1(2)=3$. Please advise.


2 Answers 2


enter image description here Going through the reflections in the order pictured, you can see that you end with what I labeled as $r_1$ (most books use $\rho$ to denote rotations. paint doesn't have greek. >_<)

Going through the reflections in the other order will give you what I labeled as $r_2$.

The point of the exercise is to show that the reflections aren't abelian because you can do things in a different order and get different results and aren't a subgroup because you can get something other than a reflection (the rotations by themselves are however an abelian subgroup).

  • 1
    $\begingroup$ Thanks a ton for putting so much effort to explain this .. you are awesome :) $\endgroup$
    – square_one
    Oct 26, 2014 at 3:44
  • $\begingroup$ You are quite welcome. Also, a fair warning, some literature refers to the group as $S_n$ to refer to the symmetric group of symmetries on $n$ elements, while others refer to the same group as $S_{2n}$ since there are $2n$ symmetries. Be careful if/when you change books to keep track of what their personal notation use is. $\endgroup$
    – JMoravitz
    Oct 26, 2014 at 3:46

Your conclusion should be that $$\phi_1 \circ \phi_3 \neq \phi_3 \circ \phi_1,$$ which is correct. The group $S_3$ is a standard example (in fact, the smallest example) of a nonabelian group, in which (by definition) not all pairs of elements commute.

As for visualizing, I like to think of $S_3$ as the group of isometries (symmetries) that fix an equilateral triangle. For example, we can label the vertices $1, 2, 3$, in which case we can identify (for example) $\phi_1$ as the reflection that fixes vertex $1$ and exchanges $2$ and $3$. In this realization, the transpositions are reflections, and the $3$-cycles are rotations.

Likewise, one can visualize $S_4$ as the group of isometries of a regular tetrahedron; after this one runs out of a easily visualizable dimensions, and so you might prefer to view $S_n$, $n > 4$, as the set of shuffles of a deck of $n$ cards.

  • $\begingroup$ Thanks .. but my doubts are at a more basic level than your explaination .. I don't see how/why $\phi_3(1)$=2 and $\phi_1$(2)=3 .. :( $\endgroup$
    – square_one
    Oct 26, 2014 at 3:30

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