If $K\lhd G$ and $(|K|,[G:K])=1$ then $K$ is characteristic. Here is my Idea I think I am somehow supposed to use the fact that $K$ is a unique subgroup of order $|K|$. To prove this. I think the proof is that for every automorphism of G ($\varphi$), $|\varphi(K)|=|K|$. But I am not sure if that is true, or if it is how to prove that it is true.
 A: Suppose $K\triangleleft G$, $(|K|,[G:K])=1$ and $H$ is a subgroup of order $|H|=|K|$.  Let $h\in H\setminus K$.
Since $h$ is not the identity, $|h|>1$ and divides $|K|=|H|$.  Now, consider the image of $h$ in $G/K$ by the natural surjection $\phi:G\rightarrow G/K$ (since $K$ is normal).  The image is not the identity since $h\not\in K$, so $|\phi(h)|>1$.  Moreover, $|\phi(h)|$ must divide $|h|$ (which divides $|K|$) as well as $|G/K|=[G:K]$.
This is impossible, because the only positive common divisor of $|K|$ and $[G:K]$ is $1$, which cannot be the order of $\phi(h)$.
A: The claim that for every automorphism $\varphi$ of $G$, $|\varphi(L)| = |L|$ is true for every subgroup $L$, because $\varphi$ is injective. That doesn't mean in general that $\varphi(L) = L$ though.
To show $K$ is the unique subgroup of order $|K|$ with the given assumptions, let $H$ be another subgroup of $G$ with order $|K|$. Since $K$ is normal, $HK$ is a subgroup and $|HK| = \frac{|H||K|}{|H \cap K|} = \frac{|K|^2}{|H \cap K|}$. Thus $|HK|$ divides $|K|^2$. But $|HK|$ also divides $|G| = |K| |G:K|$. Since $(|K|,|G:K|) = 1$, $|HK|$ must divide $|K|$ (why?). But $K$ is contained in $HK$, so $HK = K$, which means $H \subseteq K$. But $|H| = |K|$, so $H = K$.
Now you can deduce that since $\varphi(K)$ is also a subgroup of order $|K|$, we must have $\varphi(K) = K$; thus $K$ is characteristic.
A: I think I just got it. Let $\phi: G\to G$ be an automoprhism, and let $\varphi:G\to G/K$ be the natural map. Then it follows that $|\phi(K)|$ divides $|K|$ so $\varphi (\phi(K))=\{e^*\}$ so it follows that $\phi (K)\subset K$.
