I'm trying to do the following exercise from Rotman's book, An Introduction to Theory of Groups.
6.10 i) If $A,B,$ and $C$ are finite abelian groups with $A\oplus C \cong B \oplus C$, then $A \cong B$.
Hint: Use Exercise 6.9.
Exercise 6.9 says that:
6.9 If $G$ and $H$ are finite abelian groups then $$U_p(n,G\oplus H)=U_p(n,G)+U_n(p,H)$$ for all primes $p$ and $n \geq 0$.
I found the following solution on this site, but I think it is not a duplicate, as the proposed solution is different.
The problem is that $U_p(n,G)$ is only defined for $p-$primary group, that is:
Definition (page 131): If $G$ is a $p-$primary abelian group and $n \geq 0$, then $$U_p(n,G)=d(p^nG/p^{n+1}G)-d(p^{n+1}G/p^{n+2}G),$$ where $d(H)$ is the dimension of $H$ as considered as a vector space over $\mathbb{Z}_p$ if $pH=\{0\}$ for some prime $p$.
My attempt: I started considering $A, B$, and $C$ $p-$primary and I did Exercise 6.9. After that, I applied this result to my problem, getting $$U_p(n,A)+U_p(n,C)=U_p(n,A\oplus C)=U_p(n,B\oplus C)=U_p(n,B)+U_p(n,C)$$ which implies that $$U_p(n,A)=U_p(n,B) \hbox{ for all } n \geq 0. \tag{$\star$}$$ Now, Corollary $6.12$ ensures that if $G$ and $H$ are finite $p-$primary abelian groups, then $G \cong H$ if, and only if, $U_p(n,G)=U_p(n,H)$ for all $n \geq 0$. Therefore, from $(\star)$ we obtain that $A \cong B$.
