Suppose we need to prove statement of the form either $P$ or $Q$. Is it sufficient to prove $P$ whenever not $Q$. Or do we need to show $Q$ whenever not $P$ holds as well?
I was trying to prove this:
Let $G$ have order $4$. Either $G$ is cyclic or every element of $G$
is its own inverse. Thus, show that G is abelian.
Here's my own attempt of proof.
Let $|G|=4$. Suppose there is no element $x\in G$ such that $x=x^{-1}$ i.e. $x^2=e$. Then we need to show $G$ is cyclic. Let $y\in G$ with $y\ne e$ be arbitrary. Thus, $\langle y \rangle$ is a cyclic subgroup of $G$. By Lagrange's Theorem, $|\langle y\rangle |= |y|$ divides $|G|$. Since, $|y|$ cannot be equal to $2$ (by our supposition), it must be the case that $|y|=4$. Hence $G=\langle y\rangle $.
In a similar way, we could show if $G$ is not cyclic, every element of G is its own inverse. Abelianess follows from these two consequences.
Best Answer
Usually by "either P or Q" a mathematician means "P is true, or Q is true, or both". This is in contrast to the usual English interpretation, where it would more usually mean the same thing as a mathematician's "exactly one of P or Q is true".
To answer your particular question: it's enough to show that if not P, then Q; or you could instead prove that if not Q, then P. Then the contrapositive is "if not P, then Q", which is automatically true since the contrapositive is equivalent to the original statement, so you don't need to prove that again.
In fact it's the case that every four-element group is exactly one of "cyclic" or "has every element self-inverse", so the question of "do I mean 'or' or 'xor'" is moot. However, note that it's easy to show that a four-element group can't be both cyclic and have every element self-inverse. Indeed, if the group is cyclic, say it is $\{e, a, a^2, a^3\}$; then the inverse of $a$ is $a^{3}$ which is not equal to $a$.