Show isomorphism under specific conditions

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1. May 26, 2017

Math_QED

1. The problem statement, all variables and given/known data

Let $A,B$ be subgroups of a finite abelian group $G$

Show that $\langle g_1A \rangle \times \langle g_2A \rangle \cong \langle g_1,g_2 \rangle$ where $g_1,g_2 \in B$ and $A \cap B = \{e_G\}$
where $g_1 A, g_2 A \in G/A$ (which makes sense since $G$ is abelian and all the subgroups are normal)

(this statement is possibly false, I need it to understand a step in a proof)

2. Relevant equations

None
3. The attempt at a solution

Define $f: \langle g_1A \rangle \times \langle g_2A \rangle \rightarrow \langle g_1,g_2 \rangle$

by $f(g_1^{k_1}A,g_2^{k_2}A) = g_1^{k_1}g_2^{k_2}$ for integers $k,l$

First, we show that $f$ is well-defined:

Therefore, let $(g_1^{k_1}A,g_2^{k_2}A) = (g_3^{k_3}A,g_4^{k_4}A)$

Then, $g_1^{k_1}A = g_3^{k_3}A \iff g_1^{k_1}g_3^{-k_3} \in A$

Since, $g_1,g_3 \in B, g_1^{k_1}g_3^{-k_3} \in A\cap B = \{e_G\}$ Hence, $g_1^{k_1 } = g_3^{k_3}$. Completely analogue, we find $g_2^{k_2 } = g_4^{k_4}$.

Therefore, we conclude that $g_1^{k_1}g_2^{k_2} = g_3^{k_3}g_4^{k_4}$. Hence, $f$ is well-defined.

This also shows injectivity (every step we did is reversible).

It is also clear that this mapping is surjective. If we take $g_1^{k_1}g_2^{k_2} \in \langle g_1,g_2 \rangle$, then obviously $(g_1^{k_1}A,g_2^{k_2}A)$ gets mapped to that element.

Last but not least, we have to show that $f$ is a homomorphism. To show this,

$f[(g_1^{k_1}A,g_2^{k_2}A)(g_3^{k_3}A,g_4^{k_4}A)] = f(g_1^{k_1}g_3^{k_3}A,g_2^{k_2}g_4^{k_4}A)$

$= g_1^{k_1}g_3^{k_3}g_2^{k_2}g_4^{k_4} = g_1^{k_1}g_2^{k_2}g_3^{k_3}g_4^{k_4}$

$= f(g_1^{k_1}A,g_2^{k_2}A)f(g_3^{k_3}A,g_4^{k_4}A)$

The result follows $\quad \triangle$

Can someone verify whether my attempt is correct?

Last edited: May 26, 2017
2. May 26, 2017

Staff: Mentor

What do you mean by $\langle x,y \rangle$ and where are the elements of $A$ gone to? Isn't $g_1=g_2=e_G$ already a counterexample?

3. May 26, 2017

Math_QED

Group generated by x,y, which is a subgroup of G. This is part of a proof. I filtered context.

And what is the problem with $g_1 = g_2 = e_G$?

Then, we get that $\langle A \rangle \times \langle A \rangle = \{e_{G/A}\} \times \{e_{G/A}\} \cong \{e_G\}$, which seems fine.

4. May 26, 2017

Staff: Mentor

O.k., then $\langle . \rangle$ depends on the group. I thought it was meant to be $\langle gA \rangle =\langle ga\,\vert \,a \in A\rangle$.
However, in this case $g_1=a \in A-\{e_{G}\}\, , \,g_2=e_{G}$ will be a counterexample except in case $A=\{e_G\}$.

5. May 26, 2017

Math_QED

But $g_1 \in B$. If $g_1 \in A$, then automatically $g_1 = e_G$

6. May 26, 2017

Staff: Mentor

Forgotten. Seems I should take a nap. Sorry.

7. May 26, 2017

Math_QED

8. May 26, 2017

Staff: Mentor

Let's consider the first isomorphism theorem: $AB/A \cong B/(A\cap B) \cong B$ which gives us a hint where to look for the flaw. You have basically shown, that the projection followed by the multiplication onto the quotient group $G/A$ is a well-defined epimorphism. So what about injectivity? $AB/A \times AB/A \cong B \times B \ncong B$.

Say $f(gA,hA)=gh = e_G$. This means $g=h^{-1}$, but how do you get to $g \in A$?
(Unless I've missed something again.)

9. May 26, 2017

Math_QED

I see where you are going. I realised the injectivity is failing (at least my argument, since not every step is reversible). Thanks a lot for this. So probably, the statement is false.

10. May 28, 2017

Math_QED

I managed to prove injectivity (I could deduce from the proof's context that $\langle y_1A\rangle \cap \langle y_2A\rangle = \{A\}$, from which the injectivity follows.)

11. May 28, 2017

Staff: Mentor

$A=\langle (1),(12) \rangle\, , \,B=\langle (1),(13) \rangle\, , \,G=AB=\langle (1),(12),(13),(23) \rangle$
The stated isomorphism now reads $\langle (13)A \rangle \times \langle (13)A \rangle \cong B \times B \cong \langle (13),(13) \rangle = B$ respectively $f((13)A,(13)A) = (1)$, so how could $f$ be injective? $B \times B \cong G \ncong B$ by counting group elements.

12. May 28, 2017

Math_QED

Don't worry about it. There is a lot of context missing that can be used (this statement is part of the proof that every finite abelian group is the direct product of cyclic p-groups, which is a rather large proof, which is the reason I left as much context away as possible).

13. Jun 23, 2017

Math_QED

I am studying for my exam group theory right now and I reread this thread. How can your counterexample make sense if $G$ is finite and we already have established a surjection between the two finite groups? This clearly must imply that the function is injective as well.

14. Jun 23, 2017

Staff: Mentor

I think my misconception lied in $\langle g_1,g_2 \rangle$ which probably should be read as $\langle (g_1,g_2) \rangle$, which makes a great difference. My example ($g_1 = g_2$) used the fact that $B \times B \ncong B$ for finite groups. So if we generate a group by $g_1$ and $g_2$, we get $g_iA \cong B$ in the example. But if we generate it by the pairs $\{(g_1,g_2)\,\vert \,g_i \in B\}$ we get $B \times B$ and it's no counterexample. It all depends on how $\langle g_1,g_2 \rangle$ had to be understood. Usually it means "group generated by all by $g_1$ and $g_2$" as you also suggested in the definition of $f$. But I think we may not multiply the two and consider pairs instead.