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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.

ash
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  • You should be very clear about whether you want an ordinary OR or an exclusive OR (XOR), as I stated in a comment below my answer. You can't really depend on the natural English phrasing. This question and the accepted answer are highly relevant: https://math.stackexchange.com/questions/2130327/does-either-make-an-exclusive-or – Deepak Feb 10 '18 at 09:52
  • And please reproduce the actual statement you're required to prove in its original context. – Deepak Feb 10 '18 at 09:53
  • As I'm not knowledgeable about abstract algebra, I'll let someone else answer the specific question you're posing in your edit. I will delete my answer. – Deepak Feb 10 '18 at 09:59
  • @Deepak Thanks for the help, though! – ash Feb 10 '18 at 10:00
  • No worries. You might want to add in tags for abstract algebra, group theory etc. – Deepak Feb 10 '18 at 10:00
  • The question in the first paragraph belongs to logic, not to abstract algebra. It seems that you skipped logic; you need to learn it ASAP. – beroal Feb 13 '18 at 17:04

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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 $a$.

Patrick Stevens
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