3
$\begingroup$

In this note, I've read that $\mathbb S_n$ is a semidirect product of the alternating group $A_n$ by $\mathbb Z_2$. So I am trying to define a morphism $\rho: \mathbb Z_2 \to Aut(A_n)$ to show that $\mathbb S_n \cong A_n \rtimes Z_2$. I would appreciate suggestions on how could I define the morphism. Thanks in advance.

$\endgroup$
  • 2
    $\begingroup$ Whenever a group $G$ is presented as a semidirect product $H \rtimes K$, where $H$ and $K$ are subgroups of $G$ (or, after the fact, are identified with subgroups of $G$), with $H$ normal, the morphism $\rho \colon K \to \mathrm{Aut}(H)$ is always the same one. It's $K$ acting on $H$ by conjugation. In this case, you just need to find a subgroup $K$ of $S_n$ that is isomorphic to $\mathbf{Z}_2$ and has trivial intersection with $A_n$, and then let $K$ act on $A_n$ by conjugation. $\endgroup$ – Dave Sep 2 '14 at 5:46
  • $\begingroup$ Well, any subgroup of the form $\{e,(ab)\}$ is isomorphic to $\mathbb Z_2$ and is clearly disjoint from $A_n$. Now, you say to define $\rho(e)=e\tau e$, $\rho((ab))=(ab)\tau(ab)$ for all $\tau \in A_n$. I am not so sure how can I check from here that $\mathbb S_n$ is a semidirect product of $A_n$ by $\mathbb Z_2$. $\endgroup$ – user16924 Sep 2 '14 at 6:09
  • 2
    $\begingroup$ You probably mean $\rho(e)(\tau)$, etc. That's correct. Check that $(h,k) \mapsto hk$ is an isomorphism from $H \rtimes K$ to $G$. (Actually, surjectivity follows from the abstract criteria above only when you assume $G$ to be finite and $|H||K| = |G|$.) $\endgroup$ – Dave Sep 2 '14 at 6:20
4
$\begingroup$

Hint: Given $\mathrm{sign}$ the parity of a permutation, you have an exact sequence

$$1 \to A_n \to S_n \overset{\mathrm{sign}}{\longrightarrow} \mathbb{Z}_2 \to 0.$$

$\endgroup$
  • 2
    $\begingroup$ But the question seems to be mainly about the splitting of that sequence, not the existence. $\endgroup$ – Tobias Kildetoft Sep 2 '14 at 9:56
  • $\begingroup$ The OP did not seem to know how to start, and the sequence precisely suggests to take a morphism $f : \mathbb{Z}_2 \to S_n$ so that $\mathrm{sign} \circ f = \mathrm{Id}$, and define $\rho : x \mapsto (y \mapsto f(x)yf(x)^{-1})$. But it's my opinion. $\endgroup$ – Seirios Sep 2 '14 at 13:13
  • $\begingroup$ @TobiasKildetoft Can you give me some details on how to proceed to show that the sequence splits? And if it does split, doesn't that mean that $S_n \cong \mathbb{Z}_2 \oplus A_n$?? And we want to the semi direct product?? Or from that, we get that $\frac{S_n}{A_n} \cong \mathbb{Z}_2$ and so $A_n \triangleleft S_n$ which is really what this was about..? Well actually that would come from being of index 2.... So for $S_n$ to be a semi direct product of the two, it seems the 3 conditions are all pretty obviously fulfilled. Is there something I am missing? $\endgroup$ – Mathematical Mushroom Jul 10 at 22:30
3
$\begingroup$

More generally:

Theorem. If $G$ has a subgroup of index two, $H$ say, and there exists an element $g\not\in H$ of order two then $G$ splits as a semidirect product $G=H\rtimes\mathbb{Z}_2$.

This is because $\langle g\rangle\cap H=1$, and the other (internal) semidirect product conditions follow because $H$ has index two (so is normal, and so on).

So, in order to answer your question you simply need to find an element of order two in $S_n$ which is not contained in $A_n$.

$\endgroup$
0
$\begingroup$

What are the permutations in $S_n$ that are not in $A_n$? How do the operate on $A_n$. Take $S_3$ as an example.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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