# 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?

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What is $g$? An element of $G$ would be a fishy thing to say. –  user21436 Apr 6 '12 at 9:38
Is $f(g)$ a typo for $f(G)$? –  Brian M. Scott Apr 6 '12 at 9:40
@Brian Wouldn't that render this problem trivial? Not that I know of a way to make it non-trivial, : ) –  user21436 Apr 6 '12 at 9:41
@BrianM.Scott NO, f(g) is not a typo for f(G) –  faisal Apr 6 '12 at 9:50
@faisal Then what is $g$? –  user21436 Apr 6 '12 at 9:54

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.

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How did you get the second step? –  faisal Apr 6 '12 at 11:06
Note that $f:G\rightarrow G$ when restricted to $H$ gives an homomorphism $f_{|H}:H\rightarrow G$ whose kernel is $ker f_{|H}=ker f\cap H$. –  Josué Tonelli-Cueto Apr 6 '12 at 11:12
How does your last step shows that the [f(G):f(H)] divides [G:H]? –  faisal Apr 6 '12 at 11:31
When dividing the two wqualitis, you have $$|f(G):f(H)|\frac{|ker f|}{|ker f\cap H|}=\frac{|f(G)|}{|f(H)|}\frac{|ker f|}{|ker f\cap H|}=\frac{|G|}{|H|}=|G:H|$$ from where divisibility is obtained since $\frac{|ker f|}{|ker f\cap H|}$ is an integer. –  Josué Tonelli-Cueto Apr 6 '12 at 11:32
Sorry for the typo. –  Josué Tonelli-Cueto Apr 6 '12 at 11:37

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 $[G:H]$.

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