$\gcd(|G|, |\text{Aut}(G)|)=1$ means G is abelian? 
Prove the following assuming that $G$ is finite group with $\gcd(|G|, |\text{Aut}(G)|)=1$.
a) G is abelian (done).
b) Every Sylow subgroup of $G$ is cyclic of prime order.

Since G is abelian than every Sylow subgroup is unique, but does it mean cyclic?
Any suggestion?
 A: For the first part, the inner automorphism group $\text{Inn}(G)$ is a subgroup of $\text{Aut}(G)$ so $\lvert\text{Inn}(G)\rvert$ divides $\lvert\text{Aut}(G)\rvert$.
However, $\text{Inn}(G)\cong G/Z(G)$ so $\lvert\text{Inn}(G)\rvert$ divides $\lvert G\rvert$. In our case where $\gcd(\lvert\text{Aut}(G)\rvert,\lvert G\rvert)=1$, this forces $\text{Inn}(G)=1$. Since $\text{Inn}(G)\cong G/Z(G)$, we must have $G=Z(G)$, meaning that $G$ is abelian.
For the second part, we will use the classification of finite abelian groups which states that
$$G\cong C_{p_1^{a_1}}\times C_{p_2^{a_2}}\times\cdots\times C_{p_k^{a_k}}$$
for (not necessarily distinct) primes $p_j$. If some $a_j\geq2$ then
$$\lvert\text{Aut}(C_{p_j^{a_j}})\rvert=\varphi(p_j^{a_j})=p_j^{a_j-1}(p_j-1)$$
is divisible by $p_j$. However, $\text{Aut}(C_{p_j^{a_j}})$ is a subgroup of $\text{Aut}(G)$ so $\lvert\text{Aut}(G)\rvert$ is divisible by $p_j$ which is a contradiction.
Thus, each $a_j=1$. If $p_i=p_j$ for some $i\neq j$ then
$$\lvert\text{Aut}(C_{p_i}\times C_{p_j})\rvert=(p_j^2-1)(p_j^2-p_j)$$
is divisible by $p_i$. However, $\text{Aut}(C_{p_i}\times C_{p_j})$ is a subgroup of $\text{Aut}(G)$ so $\lvert\text{Aut}(G)\rvert$ is divisible by $p_j$ which is a contradiction. Thus, $p_i\neq p_j$ for all $i\neq j$.

Moreover, it is easy to see that $p_i\nmid p_j-1$ for all $i,j$, which shows that every group of order $\lvert G\rvert$ is cyclic!
https://groupprops.subwiki.org/wiki/Classification_of_cyclicity-forcing_numbers
