If two internal products are isomorphic and one of each are isomorphic then the others are isomorphic

Question

I want to show that for abelian finite group $A$, $A = G \oplus H \simeq G' \oplus H'$ and $G \simeq G'$ then $H \simeq H'$. I found the proof : I defined $K=G\cap H'$ and $K' = G' \cap H$, and I showed that if $K$ or $K' $ is zero group then it is proved hence assumed $K$ and $K'$ nonzero. Now I want to use induction on $|G|$, so I need to prove $H \oplus G/K \oplus G'/K'$ is isomorphic to $H' \oplus G/K \oplus G'/K'$. Now by some calculation I found that I need to prove only $H'/K \oplus G' \oplus G'/K' \simeq H/K' \oplus G \oplus G/K$. And I stuck. Since $G$ is isomorphic to $G'$ the only left one is other two terms aand i'm guessing it's simple too, but I'm not sure how to prove it. What am I missing?

Answer 1

This property is called "cancellation", and it does hold for all finite groups. For a short and nice proof, see here.

For finitely generated abelian groups, the cancellation property just follows by the Fundamental Theorem of Abelian Groups. In general cancellation may not hold. For more details see here, and the links given there.