I am stuck with the following proof, and hope you can help me:
Let $N$ be a non-trivial abelian normal subgroup of $G$ and let $A \le G$ such that $|N|$ and $|A|$ are coprime. Let $A$ be abelian too, and let for all $a \in A, a \ne 1$: $C_N(a) = 1$.
Show that $A$ is cyclic.
My approach: Let $N$ and $A$ be a minimal counterexample. First I want to show that $|\pi(A)| = 1$. Suppose $p,q \in \pi(A)$ are two different primes, and consider two elements $a,b \in G$ of order $p$ and $q$ respectively. Because $ab = ba$ and they have prime order for different primes we have $o(ab) = o(a)o(b)$ and so the group $B = \langle a \rangle \cdot \langle b \rangle$ with $|B| = \frac{o(a)o(b)}{1} = pq$ is cyclic. Therefore $B$ is properly contained in $A$. Further
1) $|B|$ and $|N|$ are also coprime
2) $B$ is abelian, as a subgroup of an abelian group
3) for all $b \in B, b\ne 1$: $C_N(b) = 1$
So here I am stuck, I do not see how to derive a contradiction?
Here is a summary of a proof (from Huppert's book Endliche Gruppen I). In a minimal counterexample, $A=C_p^2$ with $p$ prime. So $A$ is a union of $p+1$ cyclic groups $A_i$ of order $p$, with any two intersecting in the identity.
I will use additive notation for $N$. For any $g \in N$ and any $i$ with $1 \le i \le p+1$, $\sum_{a \in A_i}g^a$ is centralized by each $a \in A_i$, and so it must be equal to the $0$ element of $N$. Similarly $\sum_{a \in A}g^a = 0$. But, because the identity of $A$ lies in each $A_i$, we get
$$0 = \sum_{a \in A}g^a = -pg + \sum_{i=1}^{n+1} \sum_{a \in A_i}g^a = -pg.$$
So $N$ is a group of exponent $p$, contradicting $(|A|,|N|)=1$.