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

Let $S$ be some finite set. Let $Sym(S)$ be the symmetric group on $S$. Let $A$ be the subset of permutations $\sigma$ that are such that there exist $a,b,c \in S$ such that $\sigma(a) = b, \sigma (b) = c$ and $\sigma (c) =a$ and for every other element of $S$, the action of $\sigma$ is the identity. In short, let $A$ be the set of cycles of length 3.

How do you show that the subgroup generated by $A$ is not the whole of $Sym(S)$?

share|cite|improve this question
No, I'm working through Geroch's Mathematical Physics in my spare time. This is exercise 30, I think. – Seamus Aug 30 '10 at 20:43
(and weirdly, he doesn't mention parity in any of the first three chapters on groups... No wonder I was stuck. I was trying to think how a proof would work involving those concepts I had been introduced to: cosets, commutator subgroup, normal subgroups...) – Seamus Aug 31 '10 at 12:25
@IvoTerek: DO NOT EDIT OLD POSTS FOR MINOR THINGS. THAT IS BAD BEHAVIOR! – Jyrki Lahtonen Jul 12 '14 at 19:33
Alright. Sorry. – Ivo Terek Jul 12 '14 at 19:35
up vote 6 down vote accepted

Look at the sign of a permutation. If a permutation can be decomposed into a product of even number of transpositions (aka 2-cycles), we give it a sign of 1. Otherwise give it a sign of -1. It turns out that this is a well-defined notion. The group generated by the even permutations cannot cover the odd ones, thus cannot be the whole symmetric group.

P.S. The group generated by the even permutations is usually called the alternating group.

share|cite|improve this answer
I've seen the whole odd/even partition of permutations many times, and I believe it's valid, but I've never seen it proven. I've only seen it asserted. – NovaDenizen Jul 12 '14 at 18:35

Consider the action of $S_n$ on the ring of polynomials $R=\mathbb R[x_1,\dots,x_n]$ given by permutation of the variables, and consider the element $$\Delta=\prod_{1\leq i<j\leq n}(x_i-x_j)\in R.$$ Then every $3$-cycle fixes $\Delta$, so the whole subgroup of $S_n$ generated by $3$-cycles acts trivially on $\Delta$. Yet there are elements of $S_n$, like all transpositions, which do not act trivially on $\Delta$ but map it to its opposite. Therefore $3$-cycles do not generate $S_n$.

One can alternatively use the action of $S_n$ on $\mathbb R^n$, again by permutation but now of the coordinates. Then the matrix corresponding to each $3$-cycle has determinant $1$, so every element of the subgroup generated by $3$-cycles is acts on $\mathbb R^n$ through a matrix with determinant $1$. Yet there are elements in $S_n$ whose corresponding matrix is of determinant $-1$: the same conclusion as before follows.

BN: The point of doing this in this way is to avoid having to check first that the parity of elements of $S_n$ is well-defined; all answers so far depend on doing this (I wonder if it is at all side-steppable...). In fact, this can be massaged easily into becoming a proof of that very same fact.

This is a more or less canonical example of why we want to study representations of groups.

share|cite|improve this answer

Consider the parity.

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.