Can you help me find the error in this proof? I have an idea of where it is, but want to check my reasoning (at the bottom).
The proposition is: Let G be an abelian group and A is any subgroup of G. Then $G \cong A \oplus G/A $. (but this is actually false.)
The proof for this where I am trying to find the error is:
Since G is abelian, A is a normal subgroup of G. Consider the function $h:G\oplus A \rightarrow G$ defined by $h((g,a)) = ga$. This function is clearly surjective. (*That is the part I think is wrong?)
(Rest of the proof):Furthermore, since G is abelian, $h((g_1,a_1)(g_2,a_2)) = h((g_1g_2,a_1a_2)) = g_1g_2a_1a_2 = h((g_1,a_1))h((g_2,a_2))$ so $h$ is a homomorphism. By isomorphism theorems, $(G\oplus A)/Ker(h) \cong G$. Since the kernel of $h$ is the subgroup $\{(a,a^{-1}) | a \in A\}$, this kernel is a subgroup of $(G \oplus A)$ and isomorphic to A. Two cosets of kernel of h are equal when $(g_1,a_1)Ker(h) = (g_2,a_2)Ker(h)$. $(g_2^{-1}g_1, a_2^{-1}a_1) \in Ker(h)$, so $g_2^{-1}g_1a_2^{-1}a_1 = 1$ so $g_2^{-1}g_1 \in A$ and $g_2A = g_1A$. This shows that the association $\phi:G/A\oplus A \rightarrow (G\oplus A)/Ker(h)$ defined by $\phi(gA,a) = (g,a)Ker(h)$ is a function (and also homomorphism and bijection). So we conclude the proposition.
*My reasoning for thinking the function is not surjective is I'm thinking of a situation where G = ($\mathbb{R}, *) $ and A = {0}. Then the function given would not be surjective because Image($h$) $\neq G$. Am I right here or does this part of the proof hold?