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I'm having a hard time showing an equality of indices holds. This is exercise 1.44 from Lang's Algebra.

Suppose f\colon A\to A' is a homomorphism of abelian groups, and $B\leq A$. Denote by $A^f$ and $A_f$ the image and kernel of $f$ in $A$ respectively, and similarly for $B^f$ and $B_f$. Show that $[A:B]=[A^f:B^f][A_f:B_f]$ in the sense that if two of these three indices are finite, so is the third, and the stated equality holds.

My initial thought was to compose a bijection from $A/B$ to $A^f/B^f\times A_f/B_f$. I tried to map a coset $aB$ to $(f(a)B^f,\underline{\hspace{.75cm}})$ but couldn't think of a nice candidate for the right coordinate.

Does anyone have suggestions on the right approach here? Thanks for your input.

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    I left out one important detail in my suggestion: the maps aren't just injective/surjective, but we actually have isomorphisms $A^f\cong A/A_f$, and similarly in the other two cases. If we have finite groups $N\subset A$, then $|A/N|=|A|/|N|$. Thus, we need to establish an isomorphism between $A_f/B_f$ and the kernel of the map $A/B\to A^f/B^f$.2011-09-21

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This follows from the isomorphism theorems and the fact that for finite groups, $|A/B||B|=|A|$.

First, note that we have the following useful isomorphisms. $A^f=A/A_f$, $B^f=B/B_f$, $B_f=B\cap A_f$. To simplify notation, let $A_f=N$.

We have two sets of maps in the problem: $N\to A \to A/N$ and $N\cap B \to B \to B/(N\cap B)$, where the maps are the natural inclusions and quotient maps. What we wish to show is that we have maps

$N/(N\cap B)\to A/B \to (A/N)/(B/(N\cap B))$, and that $(A/B)/(N/N\cap B)\cong (A/N)/(B/(N\cap B))$. However, using the isomorphism theorems, we have

$(A/B)/(N/N\cap B)\cong (A/B)/(N+B/B)\cong A/(N+B)\cong (A/N)/((N+B)/N) \cong (A/N)/(B/N\cap B)$.

If you trace through the isomorphisms in the isomorphism theorems explicitly, you will see that they give exactly the maps you want them to.

More concretely, because $B_f=A_f\cap B$, the map $A_f/B_f\to A/B$ sending $aB_f$ to $aB$ is injective, and because the map $A\to A^f$ is surjective, the map $A/B\to A^f/B^f$ sending $aB$ to $f(a)f(B)$ is also surjective. The statement that $aB$ is in the kernel of the second map is equivalent to the statement that $f(a)=f(b)$ for some $b\in B$, or $f(ab^{-1})=e$, hence $aB\in A_f/B=\operatorname{im}(A_f/B_f)$, and hence the image of the first map is equal to the kernel of the second map.

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    Thanks for spelling it out for a simpleton like me, much appreciated.2011-09-21