# Direct product of abelian groups. Isomorphism.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
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.

#### Deveno

##### Well-known member
MHB Math Scholar
i don't have an answer, but i can tell you your approach is doomed.

let A = B = C = Z2.

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.

Staff member

MHB Math Scholar

#### caffeinemachine

##### Well-known member
MHB Math Scholar
i don't have an answer, but i can tell you your approach is doomed.

let A = B = C = Z2.

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.