Show that (A:B) = (A^f:B^f)(A_f:B_f) in Abelian Groups

[eastside00_99]

Let A and A' be abelian groups and B a subgroup of A. Let f: A -> A' be a group homomorphism. Let $A^f = Im(f), B^f = Im(f|_B), A_f = Ker(f), B_f=Ker(f|_B)$.
Show that $(A:B) = (A^f:B^f)(A_f:B_f)$.

This is the work I have done: We have $B \subset E=f^{-1}(f(B))$. We also have an isomorphism from $A/E$ to $A^f/B^f$. We have the inclusion map $\iota : A_f \rightarrow E$. We can use this to define a homomorphism $\iota_* : A_f/B_f \rightarrow E/B$ by xB_f goes to xB. This is well defined because B_f is contained in B. It is also one to one for xB=B implies x is in B and x in A_f which gives x in B_f. The last thing I need to show is surjectivity...

This just seems hopeless to me to. But, if I can show that then we have the theorem. Note, I really haven't used the fact that the groups are abelian which is almost surely cause for thinking that I'm not on the right track.

 

[alyscia]

I think just chasing the index using the first isomorphism theorem is good enough.

 

[eastside00_99]

Yeah, your right. thanks

 

[gauss^2]

Sorry to bump an old thread, but I just started reading through Lang and haven't cracked this problem either. It has been years since I have taken an algebra class (and only at undergrad level), so it could be the rust though. By chasing the index, is it meant we use the 1st iso thm to say that $A/A_f \approx A^f$ (and the same for $B$), so that we have
$(A:A_f) = (A^f:1)$ and $(B:B_f) = (B^f:1)$, where $(G:1)$ is the order of the group $G$?

So then we could say that
$(A:1) = (A:A_f)(A_f:1) = (A^f:1)(A_f:1)$.

Since $(A^f:1) = (A^f:B^f)(B^f:1)$ and $(A_f:1) = (A_f: B_f)(B_f:1)$, we then get
$(A:1) = (A^f:B^f)(A_f:B_f)(B^f:1)(B_f:1)$

as well as

$(A:1) = (A:B)(B:1) = (A:B)(B:B_f)(B_f:1) = (A:B)(B^f:1)(B_f:1)$

But that's as far as you can get if the restriction of $f$ to $B$ has infinite kernel or image. Or am I missing something really simple?

 

[micromass]

Sorry to bump an old thread, but I just started reading through Lang and haven't cracked this problem either. It has been years since I have taken an algebra class (and only at undergrad level), so it could be the rust though. By chasing the index, is it meant we use the 1st iso thm to say that $A/A_f \approx A^f$ (and the same for $B$), so that we have
$(A:A_f) = (A^f:1)$ and $(B:B_f) = (B^f:1)$, where $(G:1)$ is the order of the group $G$?

So then we could say that
$(A:1) = (A:A_f)(A_f:1) = (A^f:1)(A_f:1)$.

Since $(A^f:1) = (A^f:B^f)(B^f:1)$ and $(A_f:1) = (A_f: B_f)(B_f:1)$, we then get
$(A:1) = (A^f:B^f)(A_f:B_f)(B^f:1)(B_f:1)$

as well as

$(A:1) = (A:B)(B:1) = (A:B)(B:B_f)(B_f:1) = (A:B)(B^f:1)(B_f:1)$

But that's as far as you can get if the restriction of $f$ to $B$ has infinite kernel or image. Or am I missing something really simple?

You must only show that if two of the indices $(A:B),(A_f:B_f),(A^f,B^f)$ is finite, then so is the third. Right?

Well, maybe you can start by showing that

$$0\rightarrow A_f/B_f\rightarrow A/B\rightarrow A^f/B^f\rightarrow 0$$

is exact.

 

[gauss^2]

You must only show that if two of the indices $(A:B),(A_f:B_f),(A^f,B^f)$ is finite, then so is the third. Right?

Well, maybe you can start by showing that

$$0\rightarrow A_f/B_f\rightarrow A/B\rightarrow A^f/B^f\rightarrow 0$$

is exact.

Apologies for not getting back sooner, micromass... I appreciate the help.

So essentially, if that is an exact sequence then the homomorphism $\varphi \colon A_f/B_f \to A/B$ embeds an isomorphic copy of $A_f/B_f$ in $A/B$ since it is injective. Then since the homomorphism $\phi \colon A/B \to A^f/B^f$ is surjective with kernel equal to $\varphi(A_f/B_f) \approx A_f/B_f$, we then have $(A/B)/(A_f/B_f) \approx A^f/B^f$, correct?

I'm not sure what the two maps $\varphi \colon A_f/B_f \to A/B$ and $\phi \colon A/B \to A^f/B^f$ should be though. If I define $\varphi(aB_f) = aB$, then it is well-defined, is a group homomorphism because $B_f$ and $B$ are normal in $A$ (the latter thanks to $A$ being abelian), and has image $A_f/B$.

My first instinct is to define $\phi(aB) = f(a)B^f$. This is well-defined, is a group homomorphism since $B$ is normal in $A$ and $B_f$ is normal in $A_f$, is surjective, and has kernel $f^{-1}(B^f)/B$; not $A_f/B$.

Where am I going wrong here? Thanks.

 

[micromass]

Maybe you can prove that

$$f^{-1}(B^f)/B=A_f/B$$

??

 

[gauss^2]

Maybe you can prove that

$$f^{-1}(B^f)/B=A_f/B$$

??

Duh, I was sure they weren't equal, lol. Thanks so much for the help!

 

[gauss^2]

Just in case someone else is struggling with this problem in the future and finds this thread in google, let me show why $A_f/B = f^{-1}(B^f)/B$

1. Assume $aB \in A_f/B$ so that $f(a) = e'$, where $e'$ is the identity element of $A'$.

2. Then $e' \in B^f$ shows that $a \in f^{-1}(B^f)$, so we see that $aB \in f^{-1}(B^f)/B$ and therefore $A_f/B \subset f^{-1}(B^f)/B$ holds.

3. Let $aB \in f^{-1}(B^f)/B$ so that $f(a) \in B^f$.

4. Then there is some $b \in B$ such that $f(a) = f(b)$.

5. Then $f(a b^{-1}) = f(a)f(b)^{-1} = e'$, so $a b^{-1} \in A_f$.

6. Then $a = (a b^{-1})b$, so $aB \in A_f/B$ and therefore the quotients $A_f/B$ and $f^{-1}(B^f)/B$ are equal.