Finite Abelian groups: $G \times H \cong G\times K$ then $H\cong K$

Let $G,H,$ and $K$ be finite abelian groups. If $G \times H \cong G\times K$ then $H\cong K$.

I am trying to use the fundamental theorem for abelian groups to solve this, it is clear intuitively that I can decompose $G \times H$ and $G \times K$ and then cancel out the factors of $G$ but I have no clue on writing this rigorously, or if my method is right.

Edit: Can I say that $H \cong \Bbb{Z}_{p^i} \times \ldots \times \Bbb{Z}_{p^n}$ and $K \cong \Bbb{Z}_{p^q} \times \ldots \times \Bbb{Z}_{p^k}$ and since $G \times H \cong G\times K$ then $G\times \Bbb{Z}_{p^i} \times \ldots \times \Bbb{Z}_{p^n} \cong G \times \Bbb{Z}_{p^q} \times \ldots \times \Bbb{Z}_{p^k}$ , but the representation is unique hence $\Bbb{Z}_{p^i} \times \ldots \times \Bbb{Z}_{p^n} \cong \Bbb{Z}_{p^q} \times \ldots \times \Bbb{Z}_{p^k}$ that is $H \cong K$?

Solution 1:

Recall that the Fundamental Theorem for finitely generated Abelian groups (which applies of course to finite abelian groups) states that every finite abelian group is isomorphic to, and can be decomposed uniquely as, the direct product of cyclic groups of prime order and/or cyclic groups of order equal to the power of a prime.

So each of $G, H, K$ can be decomposed uniquely in such a manner, and so the direct product of $G\times H$ and of $G\times K$ can each thereby be expressed uniquely (up to isomorphism) as the direct product of cyclic groups.

• Since $G$ is a common factor to each of $G \times H$ and $G\times K$, and their decompositions are unique up to isomrophism,
• and given $G\times H \cong G\times K$

then what does this imply about the relationship between $H$ and $K$?