As the title says, I want to show that
Given the $\mathbb{Z}$-module $\mathbb{Z}$, we have that the image of $\operatorname{Sym}_{2}$ is $a(1 \otimes 1)$ for even $a$.
My reasoning:
I presume that by $\operatorname{Sym}_2$ they mean the map $$\operatorname{Sym}_{2}(z) = \sum_{\sigma \in S_2} \sigma z$$ for $$z \in \mathcal{T}^2(\mathbb{Z}) := \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}.$$
Now, let $z$ be arbitrary. Since $$1 \otimes 1$$ is a basis for $$\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}$$ we can write every element in $$\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}$$ as $$z = \sum_{i = 1}^{n} a_i(1 \otimes 1) = \Big(\sum_{i = 1}^{n} a_i\Big)(1 \otimes 1) = a'(1 \otimes 1)$$ where $$a' = a_1 + \ldots + a_n.$$
I presume that in this case, the action of $\sigma$ on an element in $(a \otimes b) \in \mathcal{T}^2(\mathbb{Z})$ will just permute or leave unchanged, $a,b$. It is clear that this will have no effect on the element we are looking at, hence we can say that every element $$z \in \mathcal{T}^2(\mathbb{Z})$$ will remain unchanged under permutations from $S_2$.
Hence we get $$\operatorname{Sym}_{2}(z) = \sum_{\sigma \in S_2} \sigma a'(1 \otimes 1) = 2!a'(1 \otimes 1) = a(1 \otimes 1)$$ for even $a \in \mathbb{Z}$.
Now, since this holds true for arbitrary $z$, we see that any element in the image of $$\operatorname{Sym}_{2}$$ will be on the form $$a(1 \otimes 1) = a \otimes 1 = 1 \otimes a \neq 1 \otimes 1.$$
Is my reasoning correct? If not, what is wrong?
Thanks in advance.