# isomorphism argument

I am basically wondering why the two groups I have marked in red are isomorphic. I will explain something after the picture:

Let's assume that we accept that $G\simeq Z\times Z \times Z\times Z$ as they state. Then we can say that $2G \simeq 2Z\times 2Z\times 2Z\times 2Z$? But why are the the cosets groups also isomorphic?

I mean lets assume that C is normal in A, D is normal in B, and we have that $A \simeq B$, and $C \simeq D$. Then do we have that $A/C \simeq B/D$? I tried proving this, but I was not able to. Can someone please help?

• Don't use * where you actually meant $\;\times\;\;:\;\;\Bbb Z\times\Bbb Z\;,\;\;\Bbb Z*\Bbb Z\;$ are two very different groups. – Timbuc Mar 4 '15 at 9:31

Let's look at a simpler example, which I leave to you to generalize.

Suppose that $N$ is a normal subgroup of $G$, and $K$ is a normal subgroup of $H$.

Define $\phi: G\times H \to (G/N)\times (H/K)$ by:

$\phi(g,h) = (gN,hK)$, which is clearly surjective.

It should be straightforward to verify that $\text{ker }\phi = N \times K$, so by the First Isomorphism Theorem:

$(G \times H)/(N \times K) \cong (G/N)\times (H/K)$.

There is one other issue, besides the one in the answer of @DavidWheeler.

It's not just that $2G$ is abstractly isomorphic to $2 \mathbb{Z} \times \cdots \times 2 \mathbb{Z}$.

It is that $2(\mathbb{Z} \times\cdots\times \mathbb{Z})$ is equal to $2 \mathbb{Z} \times \cdots \times 2 \mathbb{Z}$, because for any $\langle n_1,\ldots,n_r \rangle \in \mathbb{Z} \times\cdots\times \mathbb{Z}$ we have $$2 \cdot \langle n_1,\ldots,n_r\rangle = \langle 2n_1,\ldots,2n_r \rangle \in 2 \mathbb{Z} \times \cdots \times 2 \mathbb{Z}$$