Prove that for a group with even order $2k$, $k$ odd, there is a subgroup $K$ with order $k$

Question

I'm trying to understand the proof my teacher did:

Consider a subgroup $H$ of $G$. If $H$ is not contained in $A_n$, then we can say that there exists at least one permutation in $H$ that is odd (remember that $A_n$ is the group of even permutations). So, if we assign the homomorphism $\phi: H\to \{1,-1\}$ such that:

$$\phi(\sigma) = \ \ \ \ 1, \sigma \mbox{ is even}$$ $$\phi(\sigma) = -1, \sigma \mbox{ is odd}$$

then, the kernel of this homomorphism is $H\cap A_n$, because the identity of $\{1,-1\}$ is $1$, and the ones that are sent to $1$ are the even permutations. But since they're not entirely contained in $A_n$, then we have to pick the intersection of both sets. So, by the isomorphism theorem, if we have an homomorphism $\phi: H \to \{1,-1\}$, the following is true:

$$\frac{H}{\ker(\phi)}\cong \{1,-1\}$$

so we have:

$$\frac{H}{H\cap A_n}\cong \{1,-1\}\implies\\\frac{|H|}{|H\cap A_n|} = |\{1,-1\}| = 2 \implies \\ \frac{|H|}{|H\cap A_n|} = 2$$

This is all from my head, as I understand, so please correct me if something's wrong. The part that I didn't understand of the proof follows now:

Ok, so, I think that we need to find an homomorphism from $G$ to $\{1,-1\}$ such that $K$ is the kernel, and also $K$ is not contained in $A_n$. If we can find this, we have that:

$$\frac{G}{K}\cong \{1,-1\} \implies \frac{|G|}{|K|} = \frac{2K}{|K|} = 2 \implies |K| = K \implies K \mbox{ is a group of odd order}$$

I understand all above. What I don't understand is how the teacher picked up the homomorphism. It tells me to pick an isomorphism from $G$ to $S_{2k}$ and consider the subgroup $H = \phi(G) < S_{2k}$.

Then it proves that $H$ is not contained in $A_n$ in this way:

Picks an $a$ such that $o(a) = 2$. Then consider a $\phi(a)\in S_{2k}$. Now $\phi(a)$ is the permutation $\pi_a: G\to G$ defined in this way:

$\pi_a(x) = ax$

Then it decomposes $\pi_a$ in cycles so we can know its parity:

$$\pi_a = (x_1 \ \ \pi_a(x_1))\cdots (x_2 \ \ \pi_a(x_k))$$

and since there are $k$ transpositions, $\pi$ is odd. So $\pi$ is not contained in $A_n$

It then concludes that $[G:K] = 2$

Could somebody explain why it didn't choose an homomorphism from $H$ to $\{1,-1\}$ and what did happen in the proof in general?

UPDATE

What I think should be done:

If I mimic the result $1$ to the second problem, we should think about an homomorphism from $G$ to $\{1,-1\}$ such that $G\cap A_{2k}$ is the kernel. Then, we could have:

$$\frac{G}{G\cap A_{2k}}\cong \{1,-1\}\implies \frac{|G|}{|K|} = |\{1,-1\}| \implies \\ \frac{2k}{|K|} = 2 \implies |K| = k$$

where the group $K$ is $K = G\cap A_{2k}$.

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UPDATE 2 - 20/08/2015:

This question, besides having been asked in the past, did not get answered using the theorem I proved here. I would like to know how to proof works using this theorem, because my book says this theorem is useful to proving my question.

Answer 1

Your professor is saying the following. The left-action of $G$ on itself gives a "permutation representation,'' that is, an embedding $G \hookrightarrow S_{2k}$. If you are not familiar with the phrase "action" in the previous sentence, all I am doing is copying the proof of Cayley's theorem that every group is a subgroup of the symmetric group; it definitely seems like you've encountered that theorem.

You seem to be writing as though $G$ were already given with as a subgroup of $S_{2k}$; I will continue to do so, but you should be aware that we are abusing notation here.

Anyway, the next step is to argue that $G$ can't be contained in $A_{2k}$. Indeed, there exists an order $2$ element, and, since $k$ is odd, its cycle decomposition, which consists of $k$ transpositions, is odd.

Now since $G$ isn't contained in $A_{2k}$, $A_{2k} \cap G$ is a proper subgroup of $G$. What is the index of that subgroup? Well, it's bounded above by $2$, since $A_{2k}$ has index 2 in $S_{2k}$. So it's exactly $2$. (You were considering the intersection with $A_{2k}$ of arbitrary subgroups $H$ of $G$, but this is unnecessary.)