Quotient groups and isomorphisms

[tmpst]

This has kind of been bugging me fo a while. Let's say we have two groups and an isomorhism [math]\varphi: G_1 \cong G_2[/math]. If [math]H_1 \lhd G_1[/math] is a normal subgroup and we denote [math]H_2 = \varphi(H_1)[/math], then surely the quotient groups [math]G_1/H_1 \cong G_2/H_2[/math] are isomorphic as well.

 

But what if we have a different isomophism [math]\theta: H_1 \cong H_3 \lhd G_2[/math]. Now we can't draw the conclusion that [math]G_1/H_1 \cong G_2/H_3[/math]. For example [math]\mathbb{Z} \cong \mathbb{Z}[/math] and [math]2\mathbb{Z}\cong 3\mathbb{Z}[/math], but clearly [math]\mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}_2 \ncong \mathbb{Z}_3 \cong \mathbb{Z}/3\mathbb{Z}[/math].

 

When can we substitute isomorphic groups when making quotient groups? If the groups are finite then nothing can conceivably can go wrong. But what about infinite groups?

[uncool]

This has kind of been bugging me fo a while. Let's say we have two groups and an isomorhism [math]\varphi: G_1 \cong G_2[/math]. If [math]H_1 \lhd G_1[/math] is a normal subgroup and we denote [math]H_2 = \varphi(H_1)[/math], then surely the quotient groups [math]G_1/H_1 \cong G_2/H_2[/math] are isomorphic as well.

 

But what if we have a different isomophism [math]\theta: H_1 \cong H_3 \lhd G_2[/math]. Now we can't draw the conclusion that [math]G_1/H_1 \cong G_2/H_3[/math]. For example [math]\mathbb{Z} \cong \mathbb{Z}[/math] and [math]2\mathbb{Z}\cong 3\mathbb{Z}[/math], but clearly [math]\mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}_2 \ncong \mathbb{Z}_3 \cong \mathbb{Z}/3\mathbb{Z}[/math].

 

When can we substitute isomorphic groups when making quotient groups? If the groups are finite then nothing can conceivably can go wrong. But what about infinite groups?

It's not even true when taking finite groups. Let [math]G = \mathbb{Z}/4 \mathbb{Z} * \mathbb{Z}/2 \mathbb{Z}, H = \mathbb{Z}/2 \mathbb{Z}[/math]. Then we can embed [math]H[/math] in [math]G[/math] in a few different ways, and some will have quotients equal to [math](\mathbb{Z}/2 \mathbb{Z})^2[/math] and some [math]\mathbb{Z}/4 \mathbb{Z}[/math].

 

What I think you might be looking for is related to something called the Ext functor; depending on your level of knowledge, it might be a bit away. You'll see it in homological algebra.

=Uncool-

[tmpst]

It's not even true when taking finite groups. Let [math]G = \mathbb{Z}/4 \mathbb{Z} * \mathbb{Z}/2 \mathbb{Z}, H = \mathbb{Z}/2 \mathbb{Z}[/math]. Then we can embed [math]H[/math] in [math]G[/math] in a few different ways, and some will have quotients equal to [math](\mathbb{Z}/2 \mathbb{Z})^2[/math] and some [math]\mathbb{Z}/4 \mathbb{Z}[/math].

 

What I think you might be looking for is related to something called the Ext functor; depending on your level of knowledge, it might be a bit away. You'll see it in homological algebra.

=Uncool-

Ah, you'r right. Seems like I didn't think it through on the finite groups. I am familiar with homological algebra, so I will definitely look up the Ext functor. Thanks for your help!

Edited August 7, 2012 by tmpst