= Solution
Let $G$ act on its underlying set by left multiplication. The resulting homomorphism
$$
\lambda:G\longrightarrow S_G\cong S_n,
\qquad
\lambda_g(x)=gx,
$$
is injective because $\lambda_g(e)=g$. This is <Cayley theorem>. If $g\ne e$, then $gx=x$ would imply $g=e$ after right cancellation, so every nonidentity permutation in the image is fixed-point free.
If $|G|$ is even, pair every element with its inverse. Elements not equal to their inverses occur in pairs. Since the identity is self-inverse and the group has even size, there must be another self-inverse element $g\ne e$. It has order two.
A permutation is odd when its <sign homomorphism>[sign] is $-1$. If a subgroup $H\leq S_m$ contains an odd element, the restriction
$$
\operatorname{sgn}:H\to\{1,-1\}
$$
is surjective. Its kernel is the set of even elements and has index two. Each coset has the same size, so precisely half of $H$ is odd.
Now let $n=4k+2$. An element of order two in the fixed-point-free regular representation is a product of
$$
\frac n2=2k+1
$$
disjoint transpositions, so it is odd. The image $H\cong G$ therefore has an index-two normal subgroup of even permutations. Its order is $2k+1>1$, so it is nontrivial and proper. Hence
$$
\boxed{G\text{ is not simple}}.
$$
Nonabelian simple groups of even order do exist; the smallest example is the alternating group $A_5$, of order sixty.
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