MHB Direct product of abelian groups. Isomorphism.

[caffeinemachine]

Let $A,B,C$ be finite abelian groups. Assume that $A\times B\cong A\times C$. Show that $B\cong C$.

I observed that $(A\times B)/(A\times\{e\})\cong B$ and $(A\times C)/(A\times\{e\})\cong C$.

So I need to show that $(A\times B)/(A\times\{e\})\cong (A\times C)/(A\times\{e\})$.

Let $\psi:A\times B\rightarrow A\times C$ be an isomorphism.

Define $\phi:A\times B \rightarrow (A\times C)/(A\times\{e\})$ as $\phi(a,b)=(\psi(a,b))(A\times\{e\})$.

If I could show that $\ker \phi=A\times\{e\}$ then I'd be done.

For that I need $\psi(a,e)\in A\times\{e\}$ for all $a\in A$, which I am unable to show and this might not even be true.

Please help.

 

[Deveno]

i don't have an answer, but i can tell you your approach is doomed.

let A = B = C = Z_2.

we have the automorphism:

(1,0)-->(1,1)
(0,1)-->(1,0)

note that is is NOT true that the image of Z₂x{0} is Z₂x{0}, it is:

{(0,0),(0,1)}.

i feel that the assumption that G is finite abelian has to be used in some essential way, and your approach does not do that.

 

[Opalg]

I am no expert on group theory, but my feeling is that maybe you need to use the structure theorem for finite abelian groups for this problem.

 

[caffeinemachine]

I am no expert on group theory, but my feeling is that maybe you need to use the structure theorem for finite abelian groups for this problem.

I never thoroughly read he structure theorem. So I think now is the time to do that.

 

[caffeinemachine]

i don't have an answer, but i can tell you your approach is doomed.

let A = B = C = Z_2.

we have the automorphism:

(1,0)-->(1,1)
(0,1)-->(1,0)

note that is is NOT true that the image of Z₂x{0} is Z₂x{0}, it is:

{(0,0),(0,1)}.

i feel that the assumption that G is finite abelian has to be used in some essential way, and your approach does not do that.

Thank you.