Normal Subgroups, quotient groups

Question

I am working on a problem in my Algebra homework and I am having some problems. The question is: if $H_1$ and $H_2$ are normal subgroups of $G_1$ and $G_2$ respectively and we know that $G_1$ is isomorphic to $G_2$ and $G_1/H_1$ is isomorphic to $G_2/H_2$, can we conclude from this that $H_1$ is isomorphic to $H_2$?

I need to prove that is true or give a counterexample if it is false.

My first idea was to use the idea behind the proof of the 1st isomorphism theorem. For this I have considered the inclusion homomorphisms $i_1: H_1 \to G_1$ and $i_2:H_2 \to G_2$, which are injective homomorphisms. These homomorphisms admit left inverses. Then I considered the surjective homomorphisms $f_1: G_1 \to G_1/H_1$ and $f_2: G_2 \to G_2/H_2$ which admit right inverses. And the last the isomorphism $\varphi:G_1/H_1 \to G_2/H_2$ that exists by assumption.

Then, we can construct the homomorphism $\psi$ from $H_1$ to $H2$ as $$\psi = i_2^{-1}f_2\varphi f_1^{-1}i_1.$$

But, I dont know if I am tired enough to confuse these things but I think that my beautiful homomorphism takes everybody in $H_1$ to the identity in $H_2$, 'cause if $h \in H_1$, then $i_1(h)=h \in G_1$, then, $f_1^{-1}(h)=hH_1=H_1$, then $\varphi(H_1)=H_2$ and $f_2(H_2)=1$ and $i_2^{-1}(1)=1.$

I don't know if this reasoning can take me somewhere. Can anybody give me a hint to solve this problem? Any hint, any reference, any idea is gonna be welcome for me.

Thank you so much, everyone.

Related: https://math.stackexchange.com/questions/40881 – Watson Oct 10 '16 at 06:47 — Watson, Oct 10 '16 at 06:47

score 4 · Accepted Answer · answered Oct 10 '16 at 04:50

4

Consider the (abelian) group $\Bbb Z_2 \times \Bbb Z_4$ (we'll use this for $G_1$ and $G_2$), so we automatically have normal subgroups.

Let $H_1 = \{\overline{0}\} \times \Bbb Z_4$.

Let $H_2 = \Bbb Z_2 \times \langle \overline{2}\rangle$.

What are your conclusions?

answered Oct 10 '16 at 04:50

David Wheeler

15,406

we can change $\overline{2}$ just by $\overline{0}$, alright? – bttmbrcelo Oct 10 '16 at 04:52
No, because in that case $G/H_2$ has order $4$, and we want it to have order $2$. – David Wheeler Oct 10 '16 at 04:54
Hummm, thank you so much. I have a lot of dificulty in Algebra. So, in this case $\mathbb{Z_1}\times\mathbb{Z_2}/H_1 = \mathbb{Z_1}\times\mathbb{Z_2}$, am I right? – bttmbrcelo Oct 10 '16 at 05:03
It's $\Bbb Z_2 \times \Bbb Z_4/H_1$, and arguing by order alone it can be seen this is cyclic of order 2, – David Wheeler Oct 10 '16 at 10:54

Normal Subgroups, quotient groups

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