# 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?

• Consider separately the cases where $a$ has finite order or infinite order. If $a$ has finite order, then there is some smallest positive $n$ such that $a^n = e$ (the identity of $G$; I'm using multiplicative notation because $G$ is not known to be abelian). What is the kernel of $\phi$ in this case? (Hint: there are infinitely many elements in the kernel.)
– user169852
Jul 7, 2019 at 4:37
• Sorry for confusion about operation. G's commutativity is unknown. If a has finite order, kernel of 𝜙 is multiple of n?
– NMZ
Jul 7, 2019 at 4:43
• That's correct, if $a$ has order $n$, then the kernel of $\phi$ is $n\mathbb Z$.
– user169852
Jul 7, 2019 at 4:44
• Thank you for your help. Finally if a has infinite order, does kernel contain every elements of its domain?
– NMZ
Jul 7, 2019 at 4:50
• Since $\phi$ is a homomorphism, we must have $\phi(n) = a^n$ for all $n \in \mathbb Z$. If $a$ has infinite order, which values of $n$ give $a^n = e$?
– user169852
Jul 7, 2019 at 4:51

## 3 Answers

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.

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 $$\Bbb Z/\operatorname {ker}\varphi\cong\{e\}$$.

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 $$\mathbf Z/\ker\varphi=\mathbf Z/\bigl(o(a)\bigr)$$.