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Can you check next statements, and they are proofs?

Statement 1. Lets $A, A_1, A_2$ - are abelian groups and $$A = A_1\oplus A_2.$$ Then $$A/A_1=A_2.$$ Proof: $$A=\{(a_1, a_2)|a_1\in A_1,~~a_2\in A_2\}.$$ $$x = (x_1, x_2)\sim y = (y_1, y_2)\Leftrightarrow x-y\in A_1 \Leftrightarrow x_2=y_2.$$ So, homomorphism $\varphi : A/A_1\to A_2$, such that $$\varphi(a_1, a_2) = a_2,$$ is isomorphism. $\blacksquare$

Statement 2. Lets $A\supset B$ - abelian, then $$A = B\oplus A/B$$

Proof: $A\supset B$, therefore $$\exists C\subset A: A=B\oplus C.$$ And from first statement: $$C = A/B.$$ $\blacksquare$


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It's a bad sign in a proof when you have a statement with no attempt to justify it. Why does $A \supset B$ imply the existence of a $C$ such that $A=B \oplus C$? – Chris Eagle Nov 23 '11 at 18:19
It should be true, at least for finitely generated abelian groups, that any quotient will appear as a subgroup. But Prof Magidin's example shows that you can't expect this to give a direct sum decomposition. – Dylan Moreland Nov 23 '11 at 18:57
(2⊕4)/(1⊕2) ≅ 2⊕2 ≠ 4, so you have to make sure to use the right copy of A1 in A. – Jack Schmidt Nov 23 '11 at 20:43
up vote 2 down vote accepted

First statement: you don't have equality between $A/A_1$ and $A_2$, you have isomorphism.

The second statement: you cannot hope for equality in general, though you may hope for isomorphism. However, Statement 2 is false: take $A$ to be cyclic of order $4$, $B$ to be the unique cyclic subgroup of order $2$. Then $A/B$ is cyclic of order $2$, so your assertion is that the cyclic group of order $4$ (namely, $A$) is isomorphic to a direct sum of a cyclic group of order $2$ (namely $B$) and another cyclic group of order $2$ (namely, $A/B$). This is false.

The error lies in the assertion that there must exist a $C$ contained in $A$ such that $A=B\oplus C$. There is no warrant for this assertion, as you can see with the example above.

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