Let $(G, +)$ be a group and let $K = \{ (x_1, ..., x_n) \ | \ x_1 + .... + x_n = e\} \subseteq G^n$ where $e$ is the identity of $G$. Show that $(G, +)$ is abelian if and only if $K$ is a subgroup of $G^n$.
My attempted proof: Suppose that $(G, +)$ is abelian, we show that $K$ is a subgroup of $G^n$. Pick $a, b \in K$, we show that $a +_{G^n} b^{-1} \in K$ (where $+_{G^n}$ refers to the group theoretic operation on $G^n$).
Observe that the identity of $G^n$ is $e_{G^n} = (e_1, ..., e_n)$ where $e_i = e$ for $i \in \{1, ..., n\}$.
Note that $b^{-1}$ is an element in $G^n$ (since $G^n$ is a group) such that $b^{-1} +_{G^n} b = e_{G^n}$ that is $(b_1^{-1} + b_1, ..., b_n^{-1} + b_n) = (e_1, ..., e_n)$ where $e_i =e$ for $i \in \{1, ..., n\}$.
Since $e_{G^n} \in K$ trivially (because $e+ ... + e = e$) we have then $b^{-1} +_{G^n} b \in K$ hence $b_1^{-1} + b_1 + b_2^{-1} + b_2 + ... + b_n^{-1} + b_n = e$ and by abelianness of $G$ we have $b_1^{-1} +b_2^{-1} + ... + b_n^{-1} + b_1 + b_2 + ... + b_n = e$ which implies that $b_1^{-1} + b_2^{-1} + ... + b_n^{-1} +e=e$ which implies that $b_1^{-1} + b_2^{-1} + ... + b_n^{-1} = e$ hence $b^{-1} \in K$.
Now note that $a +_{G^n} b^{-1} = (a_1 + b_1^{-1}, ..., a_n + b_n^{-1})$ and observe that since $a, b^{-1} \in K$ we have $a_1 + ... + a_n = e$ and $b_1^{-1} + ... + b_n^{-1} = e$. Hence
\begin{equation} \begin{split} a_1 + b_1^{-1} + ... + a_n + b_n^{-1} &= (a_1 + ... + a_n) + (b_1^{-1} + ...+ b_n^{-1}) \ \ \ \ \text{by abelianness of $(G, +)$}\\ & = e + e \ \ \ \ \ \text{since} \ \ a, b^{-1} \in K\\ &= e \end{split} \end{equation}
therefore we have $a +_{G^n} b^{-1} \in K$ and $K < G^n$ by the subgroup test.
Conversely assume $(G, +)$ is not abelian and that $K$ is a subgroup. We show that this results in a contradiction. Pick $a, b \in K$. Now consider $a +_{G^n} b^{-1} = (a_1 + b_1^{-1}, ..., a_n + b_n^{-1})$. Now without abelianness of $(G, +)$ we can't commute elements in the summation $a_1 + b_1^{-1} + ... + a_n + b_n^{-1}$ and thus we can't conclude that $a_1 + b_1^{-1} + ... + a_n + b_n^{-1} = e$ and therefore $a +_{G^n} b^{-1} \not\in K$. But since $K$ is a subgroup we reach a contradiction since $a +_{G^n} b^{-1} \in K$. $\square$
Is the above proof correct and rigorous? Is there any way I can make the reverse direction more rigorous?
As the comments said, the statement is only true for $n\geq 2$. Assume that we have $n\geq 2$. Then I can prove the converse direction.
Conversely, suppose $G$ is not abelian. Then $x+y\neq y+x$ for some $x,y\in G$. We have $(x,-x,0,\cdots,0),(y,-y,0,\cdots,0)\in K$. (Here we used the assumption that $n\geq 2$.) Since $K$ is a subgroup of $G^{n}$, we have $$(x,-x,0,\cdots,0)+(y,-y,0,\cdots,0)=(x+y,-x-y,0,\cdots,0)\in K.$$ So $x+y-x-y=0$. Then $x+y-x=y$, hence $x+y=y+x$, contradiction.