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Let $\epsilon : S_n \rightarrow \{\pm 1\}$ take each permutation to its sign. Then $\epsilon$ is an extension of $A_n$ by $Z_2$. Moreover, setting $s(-1)=\tau \in S_n$ determines a section of $\epsilon$ if and only if $\tau$ has order 2 and is an odd permutation.

The definition of a section is,

A section, or splitting, for $f: G \rightarrow K$ is a homomorphism $s: K \rightarrow G$, such that $f \circ s$ is the identity map of K.

I just wanted to be sure...when they say "setting $s(-1)=\tau \in S_n$ determines a section", they just mean that the homorphism that takes -1 to a permutation that has order 2 and is odd...and takes 1 to a permutation that is not of order 2 and is even. Is that correct? We can only say that "s(w)=v determines a section" if there are only two elements to consider, right?

Thanks in advance

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Usually, one speaks of $G$ being an extension of $Q$ by $K$ when $Q \cong G/K$. So, $S_n$ is an extension of $\mathbb{Z}_2$ by $A_n$. –  Sammy Black Mar 19 '13 at 22:19

3 Answers 3

up vote 3 down vote accepted

You are essentially correct. A few comments:

  • A permutation that has order 2 must be odd.
  • The splitting map is a homomorphism of groups, so it must carry identity to identity. Hence, the splitting map is completely determined by where it sends the only non-identity element, $-1$.
  • The image of the splitting, $s(\{\pm 1\}) \subseteq S_n$ is a little isomorphic copy of $\mathbb Z_2$ inside $S_n$.
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Isn't the permutation $\tau = (1,2)\circ(3,4)$ even and of order 2 ? –  Joel Cohen Mar 19 '13 at 22:25
    
Yes, of course. I wrote that too fast. :-) –  Sammy Black Mar 19 '13 at 22:30
    
I was imagining a single transposition, but as you say, a product of disjoint transpositions can have either sign. –  Sammy Black Mar 19 '13 at 22:36

Well, $s:C_2\to S_n$ must also be a homomorphism, so it cannot simply take $+1\in C_2$ to just any even permutation of $S_n$, $1$ must be taken to the identity. But given any odd permutation $\tau\in S_n$ of order $2$, there is indeed a homomorphism $s_\tau:C_2\to\langle\tau\rangle\le S_n$, and ${\rm sgn}\circ s_\tau={\rm id}_{C_2}$, so this is a section of the sign homomorphism $S_n\to C_2$.

In general, you can think of a section of $f:G\to G/N$ as a copy of $G/N$ existing within $G$. They are not always present in group extensions, only split extensions.

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The point is that to define a homomorphism $s : \{-1,1\} \to S_n$ you just need to chose $s(-1)= \tau$ because we necessarily have $s(1) = \textrm{Id}$ (because $s$ is a homomorphism).

You cannot just choose any $\tau \in S_n$ for $s$ to be a homomophism, you need to have $\tau^2 = s(-1)^2 = s((-1)^2) = s(1) = \textrm{Id}$. Conversely, you can check that for any $\tau \in S_n$ such that $\tau^2 = \textrm{Id}$, you can define a homomorphism $s : \{-1,1\} \to S_n$ (by setting $s(-1) = \tau$ and $s(1) = \textrm{Id}$).

Moreover, if $s$ is a section of $\epsilon$ we also need $-1 = \epsilon(s(-1)) = \epsilon(\tau)$. So if $\tau$ is an odd permutation of order $2$. Conversely, any such permutation $\tau$ defines a section $s$ (by setting $s(-1) = \tau$ and $s(1) = \textrm{Id}$).

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