Show that a finite group of even order that has a cyclic Sylow $2-$ subgroup is not simple.
Attempt:
Since $G$ has a Sylow $2$ subgroup so $|G|=2^km$ .
I don't know how to use the fact that the Sylow $2-$ subgroup is cyclic.
Please give some hints.
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1Lemma 2 in Tobias' answer to another question. – Jyrki Lahtonen Sep 17 '17 at 18:20
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Also https://math.stackexchange.com/questions/2404906/elements-of-odd-order-form-a-subgroup-when-the-sylow-2-subgroup-is-cyclic/ which uses some more high-powered results. – Tobias Kildetoft Sep 17 '17 at 18:52
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Let $2^km=n=|G|$, we have a permutation representation $$\varphi: G \to S_n$$ which corrsponds to left multiplication by $G$. So we treat $G$ as a subgroup of $S_n$. We claim that $G$ contains an odd permutation, hence half of $G$ is even, these elements form a subgroup of index 2, so $G$ cannot be simple.
To show $G$ indeed has odd permutation, note that under $\varphi$, a generator of a cyclic 2-Sylow must be $m$ disjoint products of $2^k$-cycle, which is an odd permutation.
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