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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?

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  • $\begingroup$ 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.) $\endgroup$
    – user169852
    Jul 7, 2019 at 4:37
  • $\begingroup$ Sorry for confusion about operation. G's commutativity is unknown. If a has finite order, kernel of 𝜙 is multiple of n? $\endgroup$
    – NMZ
    Jul 7, 2019 at 4:43
  • $\begingroup$ That's correct, if $a$ has order $n$, then the kernel of $\phi$ is $n\mathbb Z$. $\endgroup$
    – user169852
    Jul 7, 2019 at 4:44
  • $\begingroup$ Thank you for your help. Finally if a has infinite order, does kernel contain every elements of its domain? $\endgroup$
    – NMZ
    Jul 7, 2019 at 4:50
  • $\begingroup$ 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 $a^n = e$? $\endgroup$
    – user169852
    Jul 7, 2019 at 4:51

3 Answers 3

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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.

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

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

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