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$G$ and $G \times G$ where $G = \Bbb Z_2 \times \Bbb Z_2 \times \Bbb Z_2 \times\cdots$

The answer says yes but I cannot figure out what homomorphism function I could use.

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@Juan Liner: Okay, that was a bit snarky. Here's one way to think about it and so slap your forehead about how obvious it all was: Think of $G$ as being a product indexed by the natural numbers, one copy for every natural number. That is the same as having $G$ be a product indexed by the even natural numbers; or as having $G$ indexed by the odd natural numbers. Just the same, right? Well, now, when you look at $G\times G$, think of the first $G$ as being indexed by the even natural numbers, and the second as being indexed by the odd natural numbers. Then $G\times G$ is indexed by... – Arturo Magidin Nov 11 '10 at 5:24

Think of $$G = \mathbb{Z_{2_1}}\times \mathbb{Z_{2_2}} \times \mathbb{Z_{2_3}} \times \mathbb{Z_{2_4}} \times \mathbb{Z_{2_5}} \times \ldots$$ and $$G \times G= (\mathbb{Z_{2_1}}\times \mathbb{Z_{2_3}} \times \mathbb{Z_{2_5}} \times \ldots) \times (\mathbb{Z_{2_2}}\times \mathbb{Z_{2_4}} \times \mathbb{Z_{2_6}} \times \ldots)$$

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if we were to index them just like how Ross did, does it mean G and G X G are the same? – Juan Liner Nov 11 '10 at 1:05
They're isomorphic, but probably not "physically" the same since the direct product construction uses a different order. But in abstract algebra, usually when things are isomorphic then they can be regarded as the same (but not if you need to access them "physically", for example by intersecting them with each other, checking whether one contains the other, etc.) – Yuval Filmus Nov 11 '10 at 1:54
@Juan: You might enjoy Hilbert's Hotel I had seen some better writeups in the past, but can't find them now – Ross Millikan Nov 11 '10 at 6:20
@Juan: to follow up Yuval. Isomorphic has a solid definition, which is satisfied here. It allows you to conclude, for example, that operations on the structures will correspond. "The same" to me does not have a clear definition, so I don't know how to answer. – Ross Millikan Nov 11 '10 at 14:11

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