How to find kernel of a homomorphism? Given $G$ is a group and there is a homomorphism $\Bbb Z → G$ given by $φ(1) = a$, what is $\ker φ$, and $\Bbb Z/\ker φ$?
Since I don't know what is identity of $G$, I cannot identify kernel.
My guess is that kernel is just $\{0\}$ or $\Bbb Z$ because $1$ generates the  whole group $\Bbb Z$ so that $φ(n) = an$. To have $0= an$, either $a$ or $n$ need to be $0$. 
However, if my guess was right, $\Bbb Z/\ker φ$ becomes $\Bbb Z$ or $\Bbb Z/\Bbb Z$, which I don't feel right. What is mistake in my guess?
 A: As $1$ is a generator of $\mathbb{Z}$, the homomorphism is completely determined by the image of $1$. By the first isomorphism theorem we get $$\mathbb{Z}/ \text{ker}(\varphi) \cong \text{im}(\varphi) = \lbrace a^n \mid n \in \mathbb{Z} \rbrace.$$ Thus everything depends on the order of the element $a$. If the order of $a$ is finite, say $n$, then the image will be a group with $n$ elements, such that the quotient also will have $n$ elements. As one knows all the subgroups of $\mathbb{Z}$ (namely $m\mathbb{Z}$ for some natural number $m$), you will know the kernel and the quotient. By the latter you will also see what happens if the order of $a$ is not finite.
A: To take an important example, what about the canonical submersion $\varphi: \Bbb Z\to \dfrac{\Bbb Z}{n\Bbb Z}$?  The kernel is $n\Bbb Z$.
In general,  the kernel of a homomorphism is a subgroup.   The only (nontrivial) subgroups of $\Bbb Z$ are $n\Bbb Z$ for some $n$. 
Note that we will have $n=\vert a\vert$, where $\varphi (1)=a$.
So, either the kernel is infinite and $\Bbb Z/\operatorname {ker}\varphi\cong\Bbb Z_n$, the kernel is trivial and $\Bbb Z/\operatorname {ker}\varphi\cong\Bbb Z$, or $\varphi $ could be the zero homomorphism, in which case $\Bbb Z/\operatorname {ker}\varphi\cong\{e\}$.
A: The image of $\varphi$ is, by definition, the (cyclic) subgroup $\langle\mkern 1mu a\mkern 1mu\rangle\subset G$ generated by $a$, and the kernel is generated by the smallest $n>0$ such that $na =0$, i.e. by the order $o(a)$. 
Thus $\mathbf Z/\ker\varphi=\mathbf Z/\bigl(o(a)\bigr)$.
