I'm having a hard time determining what this factor group looks like

[jdinatale]

Ok, so to form the factor groups, let's say G(+)H/J, you take every element of G(+)H and left multiply by elements of J. Well elements of G(+)H look like (g, h) and elements of J look like (g, e_2)...so elements of the factor group look like (g_i, h_i)(g_n, e_2) = (g_i*g_n, h_i)

Am I completely off base, or am I on the right track?

 

[Deveno]

well, no.

here is what an element of G⊕H/J looks like:

(g,h)J.

every element of the coset (g,h)J looks like this:

(gg',he₂) = (gg',h).

in other words, every element of (g,h)J has the same second coordinate, h.

so (g,h) is in (g',h')J if and only if h = h'.

so G⊕H/J = {J,(g,h₁)J,(g,h₂)J,...}

where the h_i are distinct elements of H.

use this to prove that (g,h)J → h is an isomorphism.

step 1: prove that it doesn't matter which "g" we pick (that this map is well-defined).
step 2: prove this is a homomorphism.
step 3: prove it is bijective (injective and surjective).


another way to look at this is: J = G⊕{e}. so what you are really doing is showing that:

G⊕H/G⊕{e} ≅ H.

consider the mapping p:G⊕H→H given by p(g,h) = h. IF this is a surjective homomorphism, what can you say about G/ker(p)?

 

[jdinatale]

so G⊕H/J = {J,(g,h₁)J,(g,h₂)J,...}

First, thank you very much for the well thought out helpful response.

The above part confuses me. I feel as if G⊕H/J = \{(g, h)J \text{ }|\text{ } g \in G, h \in H\}

Why did you fix what g value?

 

[Deveno]

for convenience.

that is, suppose we have (g,h) and (g',h) in G⊕H (same h).

then (g,h)(g',h)^-1 = (g,h)(g'^-1,h^-1) = (gg',e₂), which is in J.

so two pairs (g,h), (g',h) give rise to the same coset of J.

so if we want DIFFERENT (distinct) cosets, we have to pick "different" h's.

and as long as the h's are the same, we get the same coset, so we may as well just pick one element of G, g, to represent each coset (g,h)J.