Show isomorphism under specific conditions

[member 587159]

Homework Statement

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)

Homework Equations

None

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?

 

[fresh_42]

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?

 

[member 587159]

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?

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.

 

[fresh_42]

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\}##.

 

[member 587159]

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\}##.

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

 

[fresh_42]

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

Forgotten. Seems I should take a nap. Sorry.

 

[member 587159]

Forgotten. Seems I should take a nap. Sorry.

No worries. I'm already glad I get some help :)

 

[fresh_42]

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.)

 

[member 587159]

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.)

I see where you are going. I realized 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.

 

[member 587159]

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.)

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.)

 

[fresh_42]

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.)

##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.

 

[member 587159]

##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.

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).

 

[member 587159]

##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.

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.

 

[fresh_42]

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.