On simple groups $G$, where $2\mid |G|$, $4\not\mid |G|$ The (old) exam I'm looking at has the following problem:

Suppose the order of $G$ is even, but is not divisible by $4$. Prove that $G$ is not simple.

A group with $2$ elements is clearly a counter-example to that.
Are those the only counterexamples?

(If yes, how would one prove that?)
 A: It is indeed the only counterexample. Here's how you go about showing that.
Let $|G|=2n$ where $n$ is odd, and let $x\in G$ have order $2$. The permutation $G\to G$ given by multiplication (say on the left) by $x$ is an involution, and yet it fixes nothing. Thus its cycle decomposition is comprised of $n$ $2$-cycles, whence $x$ induces an odd permutation.
The set of elements of $G$ which induce even permutations forms a subgroup $H$. For any $g\in G\setminus H$, we note that $xg$ is an element of $H$, so $g\in x^{-1}H$. This means that the index of $H$ in $G$ is $2$, and and as result $H\lhd G$.
If $|G|>2$, then $H$ is a nontrivial normal subgroup of $G$.

I learned this argument in the text Finite Group Theory by Martin Isaacs, pg. 35, and I have hardly changed it here.
A: This is a standard result, and the proof -- show that the hypotheses imply that the left action of $G$ on itself does not land in the alternating subgroup $A_{|G|}$ and thus the kernel of the sign homomorphism gives an index $2$ normal subgroup -- is also standard.  But I had a hard time remembering it: I would look it up in my undergraduate algebra text every few years.
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A few months ago though I found a construction which is a bit more general but also more conceptual, and thus I think I will remember it from now on.  The idea is that whenever a group $G$ acts on a finite set $X$ we get a homomorphism $G \ra \operatorname{Sym} X$; by composing with the natural sign map $\operatorname{Sym} X \ra \{\pm 1\}$ we get a signature homomorphism $\epsilon_X: G \ra \{ \pm 1\}$.  It is interesting to try to determine $\epsilon_X$ for various natural representations: in fact, this is related to Zolotarev's proof of quadratic reciprocity and came up in some work of mine on "abstract" generalizations thereof.
If we start with the group $G$, what is the most natural $G$-set?  Certainly it is $G$ acting on itself, say from the left (doing it on the right doesn't change what happens next).  I call the associated homomorphism $\epsilon_G: G \ra \{ \pm 1\}$ the Cayley signature map of $G$.  It turns out that there is a surprising clean characterization of $\epsilon_G$:

Lemma: Let $G$ be a finite group.
a) The following are equivalent:
(i) The Cayley signature homomorphism $\epsilon_G$ is nontrivial.
(ii) The Sylow $2$-subgroups of $G$ are cyclic and nontrivial.
b) If $\epsilon_G$ is nontrivial, its kernel is the unique index $2$ subgroup of $G$.

I discovered this result for myself and was pleased with it.  I found though that part a) previously appeared in a 1979 MONTHLY note of Patrick Morton.  For a proof and for further related results, see Lemma 3 here.  Part a) is very elementary and comes from the easy identification of the cycle type of $g \cdot$ for any $g \in G$.  Part b) uses what I think must be the very first "normal complement theorem", due in this case -- appropriately enough! -- to Cayley.  (A reference to Cayley's normal complement theorem seems to be missing currently; maybe someone can help me out here...)
Now let $G$ be finite of order $n \equiv 2 \pmod{4}$.  Then its Sylow $2$-subgroups have order $2$, so must be cyclic and nontrivial.  It follows that $\epsilon_G$ is nontrivial, and its kernel is a normal, index $2$ subgroup of $G$ (of course index $2$ subgroups of any group are always normal).  So $G$ cannot be simple unless it itself has order $2$.  Moreover $\operatorname{Ker} \epsilon$ is the unique index $2$ subgroup of order $G$.
As the OP says, this problem is often asked on qualifying exams.  In fact algebra quals often ask other problems which can be solved by this technique of extracting an index $2$ subgroup from the Cayley action of $G$ on itself.  I think that the above Lemma will help with these as well...
A: Suppose G is a nonabelian group of order 2m where m is odd, and it has a subgroup A = {1,a} of order 2 which is not normal. Note that $a^2$ = 1. Then $\exists$ y $\in$ G such that 
(1) $y^{-1}ay$  =$b \neq$ a. Then
$b^2$ = $y^{-1}ayy^{-1}ay$ = 1
$a^2$ = $b^2$ so that
aa = bb $\Rightarrow$
$ab^{-1} =ba^{-1}$ $\Rightarrow$ ab= ba
Then the set {1,a,b,ab} is a group.  You can see that by inspection (multiplying together any two elements). For example bab = bba = a.
But 4 does not divide 2m, so there cannot be a subgroup of order 4 in G.  So there is no y as in (1).
