Describing the kernel of a group homomorphism Let $\phi : \mathbb Z \to \mathbb Z_{75}$ be the function defined by $\phi(n) = 27n \mod 75$, for all $n \in \mathbb Z$.
I'm trying to describe the kernel of $\phi$ as simply as possible and so far I got...
$\phi (n) = 27n\mod {75}$ $\\$
$\phi (n) = 0$ $\\$
$27n \mod 75 = 0$ $\\$
$27n = 75k$
From here i'm not sure how to finish this, i'm assuming $\mbox{Ker}( \phi ) \ne \frac{75}{27} \Bbb Z$ since that doesn't make much sense.
 A: So you are looking for $$27n \equiv 0 \pmod{75}.$$
This is same as saying
$$9n \equiv 0 \pmod{25}.$$
$9$ has an inverse $14$ mod $25$. So multiply by $14$ on both sides to get
$$14(9n) \equiv 14(0) \pmod{25} \implies n \equiv 0 \pmod{25}.$$
Thus $n \in 25\Bbb{Z}=\{25k \, | \, k \in \Bbb{Z}\}$.
Note: An intuitive way is to think that you want $27n$ to be $0$ in mod $75$. With $27n=3(9n)$, we already have a factor of $3$, so all we need is a factor of $25$ from $9n$ so that the whole thing will be a multiple of $75$. But $\gcd(9,25)=1$ so we can't anything from $9$, so $n$ has to supply the missing factor of $25$.
A: HINT
Recall the definition of the kernel of a group homomorphism (in terms of the map you have given above):
If $\phi : \mathbb{Z} \to \mathbb{Z}_{75}$, given by $\phi(n) := 27n$ mod $75$, is a group homomorphism then the kernel of $\phi$, $Ker \phi$, is a subgroup of the domain.
So for your map $\phi$, figuring out how the kernel is defined should go something like this:
$Ker \phi :=$ {$n \in \mathbb{Z}| \phi(n) = 0$} = {$n \in \mathbb{Z}| 27n$ mod $75 =0$} =...
So, what you really need to find is all the $n$'s in $\mathbb{Z}$ such that $27n$ mod $75 = 0$. Those $n$'s will be the elements of $Ker\phi$. I recommend the method given in the other answer for this post given by Anurag A.
