The order of the image under a group homomorphism Let $f$ be a homomorphism from $G$ to $G'$. 
1) If order of a subgroup of $H$ of $G$ is $n$, then order of $f(H)$ divides $n$?
2) If the order of Kernel $f$ = $n$, then $f$ is a $n\times 1$ mapping from $G$ onto $f(G)$.
Please suggest how to proceed. Any feedback would be appreciated. 
 A: 1) Hint: Use the first isomorphism theorem.
2) Hint: Show that every member of a coset x + ker(f) gets mapped to the same point. How many elements are in each coset?
A: I'll leave out some justifications, so that you can get some practice. I'm assuming that you don't have any of the usual theorems at your disposal.
I might try to do (2) first. For that, let $K = \operatorname{Ker} f$. If there is an $x \in G$ such that $f(x) = y$, you want to show that $f^{-1}(y)$ has $n$ elements. But already we have $n$ elements in the coset $xK = \{xk : k \in K\}$, and these all map to $y$ (check these two assertions). We now have to show that in fact $f^{-1}(y) = xK$. If $z \in f^{-1}(y)$, then you should be able to show, starting from the equation $f(x) = f(z)$ and using properties of group homomorphisms, that $z \in xK$.
Now for (1). We can restrict $f$ to a surjective homomorphism $g\colon H \to f(H)$. If the kernel of $g$ has order $m$, then by the above $g$ is an $m$-to-one map. Can you take it from there?
