3

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.

Mikasa
  • 67,374
  • 5
    Both the invariant factor and primary decompositions of abelian groups are unique. Does that help? – anon Nov 02 '12 at 21:24
  • Can we just choose a side and mod out $A$? If so, this will work for nonabelian groups in addition to abelian groups. – yearning4pi Nov 02 '12 at 21:49
  • @anon: Is it useful to show that the number of elements of order $n$ in $B$ and $C$ are the same? – Mikasa Nov 03 '12 at 02:20
  • What about going back to the definition of isomorphism? If we know that $\exists \phi :A\oplus B\rightarrow A\oplus C$ such that $\phi(ab)=\phi(a)\phi(b)$, $\forall a,b\in A\oplus B$, can we either use that to construct an isomorphism from $B$ to $C$ or show that $\phi$ also is already an isomorphism from $B$ to $C$? – Todd Wilcox Nov 03 '12 at 16:44
  • Given $\phi$ as in my comment above, and looking at $(0,b_1)$ and $(0,b_2)\in A\oplus B$ ($0$ being the identities in $A$, $B$, and $C$ in what follows), suppose $\phi(0,b_1)=(0,c_1)$ and $\phi(0,b_2)=(0,c_2)$, then $\phi((0,b_1)(0,b_2))=\phi(0,b_1)\phi(0,b_2)=(0,c_1)(0,c_2)=(0,(c_1)(c_2))$ and we are almost there, right? – Todd Wilcox Nov 03 '12 at 16:54
  • I'm not sure about that notation, but I think the answer is "yes". I also am thinking that a proof like that relies on a fact that might need a lemma: If $\phi$ is an isomorphism from $A$ to $B$, then $\phi(e_A)=e_B$, where $e_A$ and $e_B$ are the respective identities for $A$ and $B$. – Todd Wilcox Nov 03 '12 at 17:32
  • @ToddWilcox: I mean by that symbol, if you take $\phi$ restricted on ${0}\oplus B$? Because I feel you took the restricted function. Right? – Mikasa Nov 03 '12 at 17:40
  • 1
    Not necessarily. I don't have a firm definition in my mind of what "restricted" or "restricted function" means. I'm also worried that I used what we are trying to prove in my outline for the proof. Specifically, that an isomorphism between two direct products can be split into isomorphisms between corresponding constituents of those products. I'd love for someone more experienced than I to weigh in on my outline. Maybe I should write it up as an answer and see if it gets downvoted. – Todd Wilcox Nov 03 '12 at 17:45
  • Ok. Thanks for your time Todd. :) – Mikasa Nov 03 '12 at 18:15
  • @ToddWilcox A restriction of a function is just taking the function over some subset of its domain. Formally the restriction of $f:A\rightarrow B$ to a subset $X\subset A$ is just $\displaystyle f\left|{X}\right.:X\rightarrow f[X]$ by $\displaystyle f\left|{X}\right.(x)=f(x)$. – Alexander Gruber Nov 03 '12 at 18:34
  • I've been thinking about this every day, and my most recent thought is that my outline using the isomorphism does not seem to require that the groups in question be finite Abelian groups, so now I'm back to thinking the Fundamental Theorem of Finite Abelian Groups is the way forward, but I can't quite make it work. – Todd Wilcox Nov 09 '12 at 22:21
  • Related: https://math.stackexchange.com/questions/370824 – Watson Dec 05 '16 at 08:35

1 Answers1

2

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.

Todd Wilcox
  • 1,283