View Full Version : Let G be a finite abelian group of order mn where (m, n) = 1

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!

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.

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

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