Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

How to show that any non transitive subgroup of the symmetric group $S_n$ is up to conjugation contained in a Young subgroup $S_k\times S_{n-k}$?

Take $\mathbb Z_3$ the subgroup of $S_3$ generated by $(123)$. This is a non transitive subgroup of $S_3$, but it is contained in no $S_k\times S_{3-k}$ unless we take $k=0$ and in then what is $S_0$ ? And even if we identify $S_0\times S_{3-0}$ with $S_3$, then actually the statement is trivial as $S_3$ contains all its subgroups!!

share|cite|improve this question
The subgroup generated by $(123)$ is transitive. For a non-transitive group, think about the orbits of $\{ 1, 2, ... n \}$ under the action of the group. – Qiaochu Yuan Jul 9 '11 at 14:35
@ Qiaochu Yuan: but $\mathbb Z_3$ is non transitive in $S_4$ as no permutation can take 1 to 4. so how can we write $\mathbb Z_3$ in this case? is it to say that $\mathbb Z_3$ is contained in the young subgroup $S_3\times S_1\subset S_4$? – palio Jul 9 '11 at 15:22
Well... yes. That's it. – Qiaochu Yuan Jul 9 '11 at 15:30
@Geoff: your accounts have been merged. – Qiaochu Yuan Jul 9 '11 at 15:32
up vote 4 down vote accepted

Suppose that $H\lt S_n$ is not transitive (that is, it does not act transitively on $\{1,2,\ldots,n\}$ under the usual action). The action of $H$ partitions $\{1,2,\ldots,n\}$ into orbits, and the action being nontransitive is exactly equivalent to the statement that this partition contains more than one equivalence class. In particular, there exist subsets $S$ and $T$ of $\{1,2,\ldots,n\}$ such that:

  • $S\cup T=\{1,2,\ldots,n\}$;
  • $S\cap T = \emptyset$;
  • $S\neq\emptyset$ and $T\neq\emptyset$
  • If $\mathscr{O}\subseteq \{1,2,\ldots,n\}$ is an orbit of $H$, then either $\mathscr{O}\subseteq S$ or $\mathscr{O}\subseteq T$.

For example, you can let $\mathscr{O}_1,\ldots,\mathscr{O}_k$ be the distinct orbits of $H$, and let $S=\mathscr{O}_1$ and $T=\mathscr{O}_2\cup\cdots\cup\mathscr{O}_k$. Since the action of $H$ is non transitive, $k\gt 1$, so $S$ and $T$ satisfy the four listed conditions.

In particular, $H$ acts on $S$ (by restriction) and $H$ acts on $T$ (by restriction); if we let $|S|=k$ and $|T|=n-k$, then we can find a $\tau\in S_n$ such that $S^{\tau}=\{1,2,\ldots,k\}$ and $T^{\tau} = \{k+1,\ldots,n\}$ (where $X^{\tau}=\left\{\tau(x)\mid x\in X\right\}$). Thus, $\tau H \tau^{-1}$ acts on $\{1,2,\ldots,k\}$ and separately on $\{k+1,\ldots,n\}$; these actions (via restriction) induce the desired embedding into $S_k\times S_{n-k}$.

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.