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**Deveno** consider the mapping:

h:G*--->G given by: h(g,e_{H}) = g.

although it is very simple, we can easily prove h is an isomorphsm of G* with G.

1) h is a homomorphism:

h((g,e_{H})*(g',e_{H})) = h((gg',e_{H})) = gg' = h((g,e_{H}))h(g',e_{H}))

2) h is injective:

suppose that h((g,e_{H})) = h((g',e_{H})). this means that g = g', so (g,e_{H}) = (g',e_{H}) (both coordinates are equal).

3) h is surjective:

suppose g is ANY element of G. then (g,e_{H}) is an element of G* that h maps to g.

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it may make more sense to consider the inverse isomorphism h^{-1}:G-->G* given by: h^{-1}(g) = (g,e_{H}). if we extend the co-domain to GxH this is an example of an embedding of G in GxH

(an embedding is a monomorphism from one group into another). even though G* isn't strictly EQUAL to G, as groups they are isomorphic (the second coordinate in G* is always e_{H}, which pretty much sits around and does nothing all day).

we use embeddings all the time, and don't even think about it, here is another example:

the integers and rational numbers are two "different kinds of things". an integer is a series of 1's or -1's:

5 = 1+1+1+1+1

-4 = (-1)+(-1)+(-1)+(-1)

a rational numbers is an equivalence class of pairs of integers:

a/b = {(c,d) in ZxZ*: ad = bc}.

nevertheless, it is common to embed the integers in the rationals by the map n--> [(n,1)] (we think of the integer 5 as 5/1 = 10/2 = 15/3 =....etc.)

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with the direct product group GxH the two homomorphisms:

p_{1}(g,h) = g

p_{2}(g,h) = h

are very important: these are called PROJECTIONS.

it turns out that these projections actually DEFINE GxH:

suppose we have a third group K, with two group homomphisms φ_{1}: K-->G, and φ_{2}:K-->H.

i claim there is a UNIQUE homomorphism ψ:K-->GxH with:

p_{1}(ψ(k)) = φ_{1}(k) and

p_{2}(ψ(k)) = φ_{2}(k), for all k in K.

how should we define ψ? it would seem the obvious way would be to set ψ(k) = (φ_{1}(k),φ_{2}(k)) (convince yourself that this works, and IS a homomorphism).

in fact the properties:

p_{1}∘ψ = φ_{1}

p_{2}∘ψ = φ_{2}

force this definition on us. just as GxH is called the (direct) PRODUCT of G and H, we can call ψ the product of the homomorphisms φ_{1} and φ_{2}: ψ = φ_{1} x φ_{2}.

this is very similar to (in some ways "just the same") the projection of a point in the plane (x,y) onto either the x-axis, or the y-axis. in fact, it turns out that as (additive/abelian) groups, we have:

R^{2} ≅ R x R. the analogue of G* is in this case, the x-axis, and perhaps it should be clearer to you that the x-axis (the points in the plane (x,0)) is indeed isomorphic to the real line.