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I need some hints to prove that:

Let $A,B,C$ are finite abelian groups such that $A\oplus B\cong A\oplus C$. Prove that $B\cong C$.

I know that every finite abelian group can be written as a finite summation $\oplus\mathbb Z_{p^k}$ and not sure about omitting the same factor.However, I have seen many times that this cancellation from both sides is done in Abelian groups. Thanks for the ideas.

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    Related: https://math.stackexchange.com/questions/3708242016-12-05

1 Answers 1

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I think I've got it.

Let each $p_i$, $q_i \in \mathbb{N}$ be a power of a prime, not neccesarily distinct. $A$, $B$, $C$ are finite and Abelian which means that $A\cong \mathbb{Z}_{p_1}\oplus \mathbb{Z}_{p_2}\oplus\cdots \oplus \mathbb{Z}_{p_k}$, $B\cong \mathbb{Z}_{p_{k+1}}\oplus \mathbb{Z}_{p_{k+2}}\oplus\cdots \oplus \mathbb{Z}_{p_{k+n}}$, and $C\cong \mathbb{Z}_{q_{1}}\oplus \mathbb{Z}_{q_{2}}\oplus\cdots \oplus \mathbb{Z}_{q_{m}}$ for some $k$, $n$, $m\in \mathbb{N}$. So, $A\oplus B\cong \mathbb{Z}_{p_1}\oplus\cdots\oplus\mathbb{Z}_{p_k}\oplus\mathbb{Z}_{p_{k+1}}\oplus \cdots\oplus \mathbb{Z}_{p_{k+n}}$ $\mathbb{Z}_{p_1}\oplus\cdots\oplus\mathbb{Z}_{p_k}\oplus\mathbb{Z}_{p_{k+1}}\oplus \cdots\oplus \mathbb{Z}_{p_{k+n}}\cong A\oplus C$ $\mathbb{Z}_{p_1}\oplus\cdots\oplus\mathbb{Z}_{p_k} \oplus\mathbb{Z}_{p_{k+1}}\oplus \cdots\oplus \mathbb{Z}_{p_{k+n}}\cong \mathbb{Z}_{p_1}\oplus\cdots\oplus\mathbb{Z}_{p_k} \oplus \mathbb{Z}_{q_{1}}\oplus \cdots \oplus \mathbb{Z}_{q_{m}}$ Since the number of terms and orders of each of the terms in the products on each side of the isomorphism are unique, it must be true that $n=m$ and it is possible to re-arrange the terms on the right so that $q_j=p_{k+j}$ for $1\leq j\leq n$, $j\in \mathbb{N}$. That means, $C\cong \mathbb{Z}_{p_{k+1}}\oplus \cdots\oplus \mathbb{Z}_{p_{k+n}}\cong B$ and we are done.

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    Thanks for your attempt Todd.2012-11-12