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Let $G$ be a finite group of even order, whose $2$-Sylow subgroups are cyclic. Show that $G$ is not simple.

I was trying to use Cayley theoem, place $G$ in some $S_m$ and get a contradiction with $A_n$ being simple but I couldn't get to anything...

Moreover I'm not sure how does it help that the $2$-Sylow subgroups are cyclic?

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marked as duplicate by B. Mehta, Shaun, Alan Wang, Taroccoesbrocco, Derek Holt finite-groups Aug 11 '18 at 7:44

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Write $|G|=n=2^rm$ when $2$ does not divide $m$. Then by definition the 2-Sylow subgroups of $G$ have order $2^r$. From the first Sylow theorem you know there is at least one 2-Sylow subgroup and you know it is cyclic. That means there is an element $x \in G$ or order $2^r$. Now let's define an action of $G$ on itself but left multiplication. ($g.h=gh$ for all $g,h \in G$). The action gives us a homomorphism $\phi:G \to S_G$ by the rule $\phi(g)(h)=g.h=gh$.

Next I assume you know that there exists an isomorphism $\psi:S_G \to S_n$, so we'll take it. And finally we'll define another homomorphism $\epsilon: S_n \to \{1,-1\}$ by $\epsilon(\sigma)=sgn(\sigma)$.

And now let's define $\psi_0=\epsilon \circ \psi \circ \phi:G \to \{-1,1\}$. This is a homomorphism of course. We'll show it is onto. If $e$ is the identity of $G$ then it is easy to see that $\psi_0(e)=1$. How can we find an element which will give us $-1$? Well, that's exactly the element $x$ which is of order $2^r$. Take any element $g \in G$. It is easy to check that $(g,xg,x^2g,...,x^{2^r-1}g)$ is a cycle in the permutation $\phi(x)$. Every cycle in the permutation of $\phi(x)$ has length $2^r$ and the number of cycles is $m$ which is an odd number. The sign of a cycle of even length is $-1$ so you get that $\psi_0(x)=(-1)^m=-1$.

So we got that $\psi_0$ is an onto homomorphism and hence by the isomorphism theorem the order of $Ker(\psi_0)$ is $\frac{|G|}{|\{-1,1\}|}=\frac{|G|}{2}$. So we found a normal subgroup of $G$ which is not trivial and not $G$ itself, so $G$ is not simple. Of course we must assume that $|G| \not =2$.

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  • $\begingroup$ I think you left out of the above otherwise nice answer the explanation of why the number of cycles $\;\left(g, xg, x^2g,...,x^{2^r-1}\right)\;$ is $\;m\;$ ( an odd number...!). This is the gist of all what you wrote... $\endgroup$ – DonAntonio Aug 11 '18 at 8:33
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    $\begingroup$ Well, I think it should be clear. There are $2^rm$ elements, and the permutation has only cycles of length $2^r$. So yes, we have exactly $m$ cycles. $\endgroup$ – Mark Aug 11 '18 at 9:07

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