Please help to understand this problem.
Let $G$ be a group, $H$ an abelian group, $\phi : G \rightarrow H$ a homomorphism. Show that $C(G) \lhd \mathrm{Ker}(\phi)$
I must be misunderstanding something, because what if we have the natural (bijective) homomorphism from $\mathbb{Z}$ to $\mathbb{Z}$. Then the kernel of the homomorphism is $0$ but the commutator subgroup of $\mathbb{Z}$ is $\mathbb{Z}$.
Any help would be appreciated.
Best Answer
It seems as if you are misunderstanding the definition of the commutator. It is not the center of the group.
Many times denoted $[G,G]$, it is defined to be the subgroup generated by the set $\{xyx^{-1}y^{-1} : x, y \in G\}$, where I am using multiplicative notation.
In your example $\mathbb{Z}$, an element in the commutator will be, for example with $x = 5$ and $y = 7$: $5 + 7 - 5 - 7$. You should be able to answer PVAL's question above, and this should give an indication of how to solve the problem.