Exercise 4.2.12 (If $\pi(G)$ contains an odd permutation then $G$ has a subgroup of index $2$)

Question

Let $G$ and $\pi$ be as in the preceding exercise. Prove that if $\pi(G)$ contains an odd permutation then $G$ has a subgroup of index $2$. [Use Exercise 3 in Section 3.3.]

Richard Ganaye · Accepted Answer

I was stuck in this exercise (I don't see how to use Exercise 3.3.3). I found a solution on stackExchange given by Andrew Dudzik.
\begin{verbatim}
https://math.stackexchange.com/questions/1530867
\end{verbatim}
\begin{proof} 
Consider the map $ f = \mathrm{sgn} \circ \pi$:
$$f : G\overset{\pi} \longrightarrow S_G \overset{\mathrm{sgn}}\longrightarrow Z_2 =\{-1,1\}.$$
Then $f$ is composed of two homomorphisms, so $f$ is a homomorphism.

By hypothesis, $\pi(G)$ contains an odd permutation, thus $f(G) 
e \{1\}$, so $f(G) = Z_2$. The First Isomorphism Theorem shows that
$$G/\ker(f) \simeq Z_2.$$
Therefore $\ker(f)$ is a subgroup of $G$ of index $2$.
\end{proof}

Exercise 4.2.12 (If $\pi(G)$ contains an odd permutation then $G$ has a subgroup of index $2$)

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Exercise 4.2.12 (If $\pi(G)$ contains an odd permutation then $G$ has a subgroup of index $2$)

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