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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: $C = A/B.$ $\blacksquare$

Thanks.

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    (2⊕4)/(1⊕2) ≅ 2⊕2 ≠ 4, so you have to make sure to use the right copy of A1 in A.2011-11-23

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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.