Below are two exercises that ask the reader to evaluate proposed proofs; the reader is to find errors (if they exist). I believe I found the error in the first. I could not find, however, an error in the second. (I am looking for confirmation or a small hint.)
$\textbf{Claim 1:}$ There exists no group containing exactly two distinct elements that do not commute.
$\textbf{Proposed Proof:}$ Assume, to the contrary, that there exists a group $G$ containing exactly two distinct elements, say $x$ and $y$, that do not commute. Thus $xy \neq yx$. Since $x$ and $y$ are the only two elements of $G$ that do not commute, $x^{-1}$ and $y$ do commute. Thus $x^{-1}y=yx^{-1}$. Multiplying by $x$ on both the left and the right, we obtain $x(x^{-1}y)x=x(yx^{-1})x$. Simplifying, we have $yx=xy$. This is a contradiction. $\blacksquare$
$\textbf{Solution:}$ The error in the proof is in the following statement: "Since $x$ and $y$ are the only two elements of $G$ that do not commute, $x^{-1}$ and $y$ do commute." How do we know that $x$ is not its own inverse? In which case, $x=x^{-1}$ and, therefore, $x^{-1}$ and $y$ do not commute. Hence, the proof is incorrect.
$\textbf{Claim 2:}$ There exists no abelian group containing exactly three distinct elements $x$ such that $x^{2}=e$.
$\textbf{Proposed Proof:}$ Assume, to the contrary, that there exists an abelian group $G$ such that $x^{2}=e$ for exactly three distinct elements $x$ of $G$. Certainly, $e^{2}=e$, so there are two non-identity elements $a$ and $b$ such that $a^{2}=b^{2}=e$. Observe that $(ab)^{2}=a^{2}b^{2}=ee=e$. Hence either $ab=a$, $ab=b$, or $ab=e$, which implies, respectively, that $b=e$, $a=e$, or $a=b,$ producing a contradiction. $\blacksquare$
$\textbf{Solution:}$ The proof is correct.
For the second, note that for groups in general, $(ab)^2 = abab \neq a^2 b^2$.
However, for abelian groups, it is true that $(ab)^2 = a^2b^2$, so that move in the proof is valid. However, it should say why it is valid (since G is abelian). Since the proof depends on the commutativity of $G$, it should say so. One such disclaimer might look like this, e.g.,