Because saying you could have $a=e$, or you could have $a^2=e$, or you could have $a^3=e$. So the order could be $1$, $2$, or $3$ (not $4$, not $5$), or $6$. This is why the definition of ORDER says smallest (or least).
So if you have an actual element $a$ of an actual group and you say its order is $6$, to show this you MUST show that orders $1$, $2$, and $3$ are ruled out. You must show that $6$ is the smallest power $n$ that gives you $a^n=e$.
So all the non-identity elements have order $2$, because each them satisfies $a^2=e$ and $a^1\ne e$. So $2$ is the SMALLEST positive integer that works.
I think your comment is perfectly reasonable, to the point and, in fact, polite. Certainly I have put much much stronger. Thankfully there are some attentive moderators who earn their title.
It always annoyed me that students criticized my books because I did not have a worked example like every possible homework question ... that's their criterion for a "good" math book.
No wonder students don't learn anything in their courses.
Well, not quite there yet. Daughter is 2nd year college in the UK, son is struggling (along with me trying to help) on his application essays. Much different process than applying to college in Ireland.
But they have a reasonable ethic. And they have survived me as a father so far.
Well, actually, my trick to teaching abstract algebra was to be very concrete and make students explore lots of examples. Try to discover the understanding/proof by understanding the right examples.
The trick to learning abstract algebra is to rush through everything as abstractly as possible and then kick yourself for thinking that was a good idea
One of my friends dads has a math phd and he gave me a copy of his thesis when I was in like middle school, cause I thought it was cool. Now that I've been actually studying math, I sometimes pick it up again and notice that I understand a couple more words in it as well.
Is it the case that $e^{f(i)n} = cos(n) + f(i)sin(n)$? I have a homogeneous linear difference equation whose solution is $y = A(-i)^n + B(i)^n$, and I'm wondering if I can arrive at the real solution by writing it as $y = Ae^{ln(-i)n} + Be^{ln(i)n} \implies y = A(cos(n) + ln(-i)sin(n)) + B(cos(n) + ln(i)sin(n)) \implies y = A cos(n) - A \frac{i\pi}{2}sin(n) + B cos(n) + B \frac{i\pi}{2} sin(n)$. The solution is supposed to be $y = C(-1)^{n + 1}$. Not sure where to go from here.
