# $ker(\phi) \cong G/im(\phi)$?

I'm currently studying up on homomorphisms and I stumbled upon a problem that I've had some trouble coming up with a proof for. In the problem we have $G$ being a finite abelian goup and $n \in \mathbb{N}$. We then define a map $\phi:G \to G$ by $\phi(g) = ng$, for all $g\in G$. The problems asks to show that $ker(\phi) \cong G/im(\phi)$.

The case where $n$ does not divide $|G|$ is straight forward since in that case we know the only element that will map to the identity is the identity since the order of every element of $G$ must divide $|G|$. Thus our kernel is trivial and hence the map is one-to-one which implies the map is also surjective and thus $im(\phi) = G$. Thus in this case it is easy to see that $ker(\phi) \cong G/im(\phi)$.

I am having a problem when $n$ divides $|G|$. The kernel should be all elements whose order is equal to $n$ and the identity but that is where I am getting stuck. I don't know how to proceed. I would be appreciative of any suggestions.

Thanks.

• $n$ not dividing $|G|$ does not mean that $ord(g)$ does not divide $n$. We might still have a non-trivial kernel. For example, $G=\mathbb{Z}_6$, $n=4$. In this case $3\in \ker(\phi)$ – Dan Jan 4 '18 at 18:06
• Oh I see that now. I should of thought about a case like that. – user1234 Jan 4 '18 at 18:16
• Is the group operation, addition? – Abishanka Saha Jan 9 '18 at 4:29

## 1 Answer

Hint: put $d= \frac{|G|}{\gcd (|G|, n)}$. What can you say about $x\mapsto dx$ ?

• I'm not exactly sure. I had a couple ideas when I tried some different cases of $G = \mathbb{Z}_k$ and $n$ but when I considered $\mathbb{Z}_2 \times \mathbb{Z}_2$ most of those ideas didn't work. – user1234 Jan 5 '18 at 16:48
• Do you see that it's a morphism ? Can you determine its image and its kernel ? – Max Jan 5 '18 at 16:51
• In my examples the kernel appears to be elements whose order divides $gcd(|G|,n)$. The image of the map I am not as certain about. – user1234 Jan 5 '18 at 16:58
• Are you certain of that ? Imagine $n$ and $|G|$ are coprime, so that $d= |G|$ : do you think the kernel is $\{0\}$ ? In fact I'm surprised you'd even find an example where this is the case – Max Jan 5 '18 at 17:30
• I see what you mean. If $d = |G|$ then our kernel would be all of $G$. – user1234 Jan 5 '18 at 17:35