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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
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1 Answer

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
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