Groups of order $pqr$ Let $G$ be a group of order $pqr$ (distinct primes) and $\varphi:G\to G$ with $\varphi(G)$ isomorphic to $\mathbb{Z}/pq\mathbb{Z}$.
Does this make $G$ abelian/cyclic? 
Is the kernel of $\varphi$ a subset of the center of $\mathbb{Z}$?
 A: It follows that $\ker(\varphi)$ is isomorphic to $\mathbb Z/r\mathbb Z$, and it's automatically normal. The homomorphism $\varphi: G \to \varphi(G) \simeq \mathbb Z/pq\mathbb Z$ selects a complementary subgroup to $\ker(\varphi)$, i.e., we have $$\ker(\varphi) \varphi(G) = G$$ and $$\ker(\varphi) \cap \varphi(G) = \{1\}.$$
(Exercise: prove this. Note that $\ker(\varphi) \varphi(G)$ is a subgroup because $\ker(\varphi)$ is normal.)
These are the building blocks of a semidirect product, so we have
$$G \simeq \mathbb Z/r\mathbb Z \rtimes \mathbb Z/pq\mathbb Z.$$
If you haven't seen semidirect products before, suffice to say there are many examples of this situation where $G$ is not commutative, and $\ker\varphi$ is not central. If you have seen semidirect products, you should be able to construct some examples.
Edit: For an explicit counterexample that doesn't require you know about semidirect products, consider the group $G = \mathbb Z/5\mathbb Z \times S_3$, where $S_3$ is the symmetry group on $3$ letters. This will form a counterexample with $p = 2, q = 5, r=3$. To see this in detail, we use the homomorphism $sgn: S_3 \to \mathbb Z/2\mathbb Z$, given by $sgn(1) = sgn( (123)) = sgn((132))=0$ and $sgn( (12)) = sgn((13))=sgn((23))=1$. This gives us a homomorphism $$id \times sgn: G \to \mathbb Z/5\mathbb Z \times \mathbb Z/2\mathbb Z,$$ $$(\overline x, \sigma) \mapsto (\overline x, sgn(\sigma)).$$ To convert this into the desired homomorphism $\varphi: G \to G$, simply observe that $\mathbb Z/5\mathbb Z \times \mathbb Z/2\mathbb Z \simeq \mathbb Z/10\mathbb Z$ and $S_3$ contains a copy of $\mathbb Z/2\mathbb Z$ (three copies, actually - one generated by $(12)$, one generated by $(13)$, and one generated by $(23)$).
A: This is not true. Consider the following group of matrices over $\mathbb{Z}/7\mathbb{Z}$:
$$
G=
\begin{Bmatrix}
\begin{bmatrix}
a & b\\
0 & 1
\end{bmatrix}\colon a,b\in \mathbb{Z}/7\mathbb{Z}, a\neq 0
\end{Bmatrix}.
$$
Then $|G|=6.7=2.3.7$. Define $\varphi\colon G\rightarrow G$ by 
$$
\begin{bmatrix}
a & b\\
0 & 1
\end{bmatrix}
\mapsto 
\begin{bmatrix}
a & 0\\
0 & 1
\end{bmatrix}.
$$
Then $\varphi$ is a homomorphism, with 
$$
\ker\varphi=
\begin{Bmatrix}
\begin{bmatrix}
1 & b\\
0 & 1
\end{bmatrix}\colon b\in \mathbb{Z}/7\mathbb{Z}
\end{Bmatrix}\cong \mathbb{Z}/7\mathbb{Z}
$$ 
and
$$Im(\varphi)
=
\begin{Bmatrix}
\begin{bmatrix}
a & b\\
0 & 1
\end{bmatrix}\colon a\in \mathbb{Z}/7\mathbb{Z}, a\neq 0
\end{Bmatrix}\cong \mathbb{Z}/(2.3)\mathbb{Z}.
$$
The element 
$
\begin{bmatrix}
1 & 1\\
0 & 1
\end{bmatrix}
$ is in kernel, but not commutes with $
\begin{bmatrix}
2 & 0\\
0 & 1
\end{bmatrix}$, hence non-central, and $G$ is non-abelian.
