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Exercise:
Be $n, min mathbb{N}$ natural numbers. We assume that $m$ is a divisor of $n$
,that means $n = m * k$ with a natural number $k$. Give an injective
Group homomorphism $f: (mathbb{Z} / mmathbb{Z}, +) rightarrow (mathbb{Z} / nmathbb{Z}, +)$. Conclude that
Isomorphic to a subset of $mathbb{Z} / nmathbb{Z}$ (i.e., there is a bijective)
Group homomorphism from $mathbb{Z} / mmathbb{Z}$ to a subgroup of $mathbb{Z} / nmathbb{Z}$.

Let $n, m$ as before. Give a surjective group homomorphism
$f: (mathbb{Z} / nmathbb{Z}, +) rightarrow (mathbb{Z} / mmathbb{Z}, +)$ and describe its core! How many
Elements has it?

Can i use $(mathbb{Z} / 3mathbb{Z},+) rightarrow (mathbb{Z} / 6mathbb{Z},+)$ and $a rightarrow a$ ?

i dont know how to continue

Let us denote $;Bbb Z_n:=Bbb Z/nBbb Z:=left{,0,1,2,...,n-1,right};$ , when "the numbers" here are in fact the corresponding equivalence classes of residues modulo $;n;$ (maybe you're used to write them as $;overline 0,,overline1,ldots$ etc.)

Define

$$phi:Bbb Z_mtoBbb Z_n;,;;phi(1):= kpmod n$$

and extend the definition naturally, meaning: for any

$$;rinBbb Z_m;,;;r=overbrace{1+1+ldots+1}^{r;text{times}}implies phi(r)=rcdotphi(1)mod n=rkpmod n;$$

(1) Prove the above is a homomorphism of groups

(2) Prove that $;kerphi=0pmod m;$

Now, you try the other direction...

Suppose $n=mk$. We can define a homomorphism
$$
fcolonmathbb{Z}tomathbb{Z}/nmathbb{Z},qquad
f(x)=k[x]_n
$$
where $[x]_n$ denotes the residue class of $x$ modulo $n$. What's the kernel of $f$? We have $xinker f$ if and only if $k[x]_n=[0]_n$, that is, $nmid kx$. This is equivalent to $mmid x$, so the kernel is $mmathbb{Z}$.

Then, by the homomorphism theorem, $f$ induces an injective homomorphism
$$
bar{f}colonmathbb{Z}/mmathbb{Z}tomathbb{Z}/nmathbb{Z}
$$
where $bar{f}([x]_m)=k[x]_n$.

Consider instead the surjective homomorphism $gcolonmathbb{Z}tomathbb{Z}/mmathbb{Z}$ defined by $g(x)=[x]_m$. Then, as $mmid n$, it is clear that $nmathbb{Z}subseteqker g=mmathbb{Z}$. Then, by the homomorphism theorem, $g$ induces a surjective homomorphism
$$
bar{g}colonmathbb{Z}/nmathbb{Z}tomathbb{Z}/mmathbb{Z},
qquad
bar{g}([x]_n)=[x]_m
$$
The kernel of $bar{g}$ is $mmathbb{Z}/nmathbb{Z}$. You can count the number of elements with Lagrange's theorem.