The index of a subgroup is divisible by the index of its image under a homomorphism Let $f$ be a homomorphism defined on a finite group $G$, and let $H$ is the subgroup of $G$. Then show that $$ \left [f(G) : f(H)\right] \text{ divides } \left [G : H\right].$$  
I know $$\left [G : H\right] = o(G)/o(H);$$
if $o(H) = n$ then $o(G) = kn$ so $o(G)/o(H) = k$.
Likewise  $$ \left [f(G) : f(H)\right] = o(f(G)) / o(f(H)).$$
I am stuck here. 
Is this the right way of doing this problem?
 A: You are near, you only need to apply the first isomorphism theorem, and deduce from it that
$$ |f(G)|\cdot|\ker f|=|G|$$
And
$$ |f(H)|\cdot |\ker f\cap H|=|H|$$
Then, you have only to notice that $\ker f\cap H$ is a subgroup of $\ker f$, and therefore
$$\frac{|\ker f|}{|\ker f\cap H|}$$
is an integer by Lagrange's theorem. Thus divisibility is obtained.
A: We can show something a little more general. Let $G$ and $K$ be two groups, and let $\phi:G\rightarrow K$ be a surjective homomorphism. If $H$ is a subgroup of $G$ with finite index $n$, then $\phi(H)$ has finite index in $K$ and its index divides $n$.
Consider the left cosets of $\phi(H)$. Since $\phi$ is surjective, each coset of $\phi(H)$ is of the form $\phi(x)\phi(H)$ for some $x\in G$. But $\phi(x)\phi(H)=\phi(xH)$ as subsets of $K$. Thus the cosets of $\phi(H)$ are images of cosets of $H$ under $\phi$. This can be interpreted as $\phi$ inducing a surjective map from the cosets of $H$ onto the cosets of $\phi(H)$. Let's give this function a name $\phi^*$. To be explicit, $\phi^*$ is the function from the set of cosets of $H$ onto the cosets of $\phi(H)$ which sends $xH$ to $\phi(xH)=\phi(x)\phi(H)$.

In general, if $X$ is a finite set, and there is a surjective set function from $X$ onto a set $Y$, then $Y$ is also finite.

Thus there only finitely many cosets of $\phi(H)$ and is thus of finite index in $K$.
Now we need to show that $[K:\phi(H)]$ divides $[G:H]$. This essentially follows from the fact that $\phi^*$ dishes out the cosets of $H$ evenly onto the cosets of $\phi(H)$.

In general, if $X$ and $Y$ are two finite sets, and $f:X\rightarrow Y$ is a surjective set function such that $f^{-1}(y_1)$ is equinumerous with $f^{-1}(y_2)$ for any $y_1$ and $y_2\in Y$, then the cardinality of $Y$ divides the cardinality of $X$.

Now we need to show that this lemma can be applied to our case here. That is, we need to show that $(\phi^*)^{-1}(C_1)$ is equinumerous to $(\phi^*)^{-1}(C_2)$ for any two cosets $C_1$ and $C_2$ of $\phi(H)$. It suffices to show that $(\phi^*)^{-1}(C)$ is equinumerous to $(\phi^*)^{-1}(\phi(H))$ for each coset $C$ of $\phi(H)$.
But what is $(\phi^*)^{-1}(\phi(H))$? With all the $\phi$s it can be difficult to untangle what it is. $(\phi^*)^{-1}(\phi(H))$ is the set of all cosets of $H$ of the form $xH$ where $x\in\ker \phi$. This is because $\phi(xH)=\phi(x)\phi(H)=e_K\phi(H)=\phi(H)$. Thus $\phi^*(xH)=\phi(H)$ and $xH\in (\phi^*)^{-1}(\phi(H))$. This is one inclusion, you can show the other.
Now let $C$ be a coset of $\phi(H)$. Then we have $C=\phi(x)\phi(H)$ for some $x\in G$. We have that $(\phi^*)^{-1}(C)$ is the set of all cosets of $H$ of the form $xyH$ where $y\in\ker\phi$. This set is in bijection with the set $(\phi^*)^{-1}(\phi(H))$ which we determined was the set of all cosets of $H$ of the form $yH$ with $y\in\ker H$. And the bijection is given by sending the coset $yH$ to the coset $xyH$, i.e., multiply each coset on the left by $x$. It's inverse is given by multiplying on the left by $x^{-1}$.
We have shown what we needed to use the lemma, thus $[K:\phi(H)]$ divides $[G:H]$.
