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?)
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 $|G|&gt;2$, 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.
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).