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Anood
04-03-2007, 10:45 AM
Let G be a finite abelian group of order mn, where (m, n) = 1. Define Gm = {g in G : order (g) | m} and Gn = {h in G : order (h) | n}.

Prove that G is ismorphic to Gm X Gn.

I think a function f has to be defined such that f:G->Gm X Gn and show that is satisfies homorphism, but I don't know how to do that. And then i think i need to show that's a bijection.

Help, please! Thank you!

pka
04-03-2007, 05:16 PM
Can you show that each group member of G can be uniquely represented as a product of some member of G<SUB>m</SUB> and some member of G<SUB>n</SUB>? Recall that if x is in G then o(x)|o(G).

Use the mapping \phi (x) = \left\{ {\begin{array}{lr}
{(x,e)} & {x \in G_m } \\
{(e,x)} & {x \in G_n } \\
\end{array}} \right.

Anood
04-03-2007, 05:35 PM
Can you clarify it more please

stapel
04-03-2007, 06:31 PM
Can you clarify it more please
You've been provided with a mapping and a suggested avenue of attack. How far have you gotten? Which bit do you need clarified?

Please be specific. Thank you.

Eliz.

pka
04-03-2007, 06:45 PM
At least show us this part.
Show that each group member of G can be uniquely represented as a product of some member of Gm and some member of Gn. Recall that if x is in G then o(x)|o(G).

Anood
04-04-2007, 02:42 PM
g^(sn) is in Gm because g^(sn)^m = g^(snm) = (g^(mn))^s = e and so
the order of g^(sn) divides m. Similarly, g^(rm)^n = e and so g^rm is in
Gn

Now how can this be used to prove homorphism and bijection