A group isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3$

Question

I am taking a first course on group theory.

I understand why $\mathbb{Z}_2\times\mathbb{Z}_3 \cong \mathbb{Z}_6$.

How can I use this fact to show that $\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_3 \cong \mathbb{Z}_2\times\mathbb{Z}_6$ ?

More in general, it looks like (but I am not sure if this is true) that given $\mathbb{G}_1\times\mathbb{G}_2\times\mathbb{G}_3$, I am allowed to take the right most "pair" (i.e. $\mathbb{G}_2\times\mathbb{G}_3$) and replace it with a group isomorphic to it. A sort of associativity rule for the external direct product.

Answer 1

The isomorphisms are given by:

$$\begin{align} \varphi: G\times (H\times K)&\to G\times H\times K,\\ (g, (h,k))&\mapsto (g,h,k);\\ \psi:G\times H\times K&\to (G\times H)\times K,\\ (g,h,k)&\mapsto ((g,h),k);\\ \theta:(G\times H)\times K&\to G\times (H\times K),\\ ((g,h),k)&\mapsto (g,(h,k)). \end{align}$$

Note that the composition of isomorphisms is an isomorphism.

Direct products are commutative via $\rho:(g,h)\mapsto (h,g)$.

Given $\pi: M\cong L$, we have

$$P\times M\cong P\times L$$

via $\pi':(p, m)\mapsto (p, \pi(m))$.