We were asked to prove the following today:
Let $G$ be a group. Prove that if $G$ has order $2n$ for some odd integer $n$ greater than $1$, then $G$ contains a proper non-trivial normal subgroup and cannot be a simple group.
We were told that Cayley's Theorem would be very beneficial to the proof but all I managed to get was to say that with a group $G$, we have a map $\varphi$ such that $$\varphi (g) = T_g $$ where $T_g$ is simply a function $G \to G$ where $T_g(x) = gx \ \ \ \forall x \in G$
Next I am told we need to show that $G$ definitely has an element of some order. Naturally I wanted to used LaGrange for this but I was told this isn't the correct approach. Searching for help after this I see that Cauchy's Theorem is used but we haven't covered that in class so we can't use that. Is there a way that I can continue from here using only Cayley's Theorem? As for a bit of reference, the exact proof of Cayley's Theorem we went over is "Every group is isomorphic to a group of permutations"
After proving we were supposed to use it to show that a group of order $216$ cannot be simple. Assuming the theorem is true, I still don't know how this can be, seeing that $216 = 2*108$. $108$ is an even number so does the theorem we are supposed to prove even play a role in this case?
This has been bugging me all morning and I'm not getting any further along looking at it by myself. So any help would greatly be appreciated.
Once you know that $G$ has an element of order $2$, the answer to the (near) duplicate question applies.
Suppose all elements of $G$ are of odd order; then the map $$f:\ G\ \longrightarrow\ G:\ g\ \longmapsto\ g^{-1},$$ has only one fixed point, which is $e$. Because $f(f(g))=g$ for all $g\in G$, the rest of $G$ is partitioned into two sets of equal size, say $m$. It follows that $|G|=2m+1$, contradicting the assumption that $G$ is even. Hence $G$ has an element of even order, say $g$ of order $2k$. Then $g^k$ is an element of order two.