Question about relationships between images and kernels of a group I have been having a reoccuring problem in my abstract algebra class with my professor defining notation and using things different from the books for homework and it's giving me difficulty to follow. I understand what the definition is saying below, but I don't quite understand anything else. 
Let $G$ be an abelian group and n be an integer. Deﬁne the map $\phi_n\colon G \to G$, $\phi_n(g) = g^n$ since $G$ is abelian $(hg)^n = h^ng^n$ that is $\phi_n$ is a homomorphism. We then have the subgroups $K_n = \operatorname{ker}(\phi_n)$ and $I_n = \operatorname{im}(\phi_n)$. So in other words, we have a map for each integer $n$ that sets up a map from the group to itself. This map 'multiplies' the element by itself $n$ times. The kernel of the map is the maps is the elements in the domain that gives you the identity. Is this correct?
Now I have to solve a series a questions using this stuff. 
Again $r$, $s$ are relatively prime. Show that $G = I_r \cdot I_s$. Hint both of these parts follow from writing $1 = ar + bs$ (I don't understand the relation at all)
Let $r$ and $s$ be relatively prime. Show that $K_r \le I_s$.
Let $\left|G\right| = rs$. Here $r$ are not necessarily relatively prime. Show that $I_s \le K_r$.
Use the preceding parts to show that if $\left|G\right| = rs$ and $r$ and $s$ are relatively prime then $I_s = K_r$.
Can someone give me a nudge in the right directions or a way to see the problem let allows me to solve all these questions? I read my book and I understand it but I don't understand any of this,
Thanks in advance
 A: For the first part, when $r$, $s$ are relatively prime, we have $1=ar+bs$ for some integers $a$, $b$ by Bézout's identity. Hence, given any $g\in G$, we may express it as
$$ g = g^1 = g^{ar+bs} = g^{ar}g^{bs} = \left(g^a\right)^r\left(g^b\right)^s = \phi_r\left(g^a\right) \phi_s\left(g^b\right).$$
From this you can conclude $G=I_r\cdot I_s$.
Most of the other questions can be solved using Bézout's identity as well, see where you get using it and comment if you need further help.

Let's show that $K_r \le I_s$ provided that $r$, $s$ are relatively prime. Take an element $g\in K_r$, so we have $g^r = e$, the identity element of the group. We need to show that $g\in I_s$. By Bezout's identity we have $1=ar+bs$ for some integers $a$, $b$. Hence,
$$
g = g^1 = g^{ar+bs} = g^{ar} g^{bs} = {\underbrace{\left(g^r\right)}_e}^a g^{bs} = g^{bs} = \phi_s(g^b).
$$
A: First of all I am guessing you have typo in the first problem? You want $G=I_r + I_s$ otherwise you are claiming the $I_r=I_s=G$ which is certainly not always true. You can't even have $I_r\cup I_s$. An easy counterexample here is $\mathbb{Z}_{12}$ with $\phi_4$ and $\phi_3$ you get $I_4=\{0,4,8\}$ and $I_3=\{0,3,6,9\}$ so you are missing $\{2,5,7,11\}$.
As for Hints:
For the first problem think of an element $g\in G$ and its images in $I_r$ and $I_s$ ($rg$ and $sg$). Without giving too much away let me point out that you would want to us the extended Euclidean algorithm to get $1g$. I might be giving too much away anyhow.
