Lately I've been trying to follow along with this YouTube lecture series called "Visual Group Theory" by Clemson's Mathew Macauley and have had some issues grasping the concept of a group's "order"..
I (think I) understand how it works when referring to reflections, rotations, etc. in the sense that we are looking for the $"k"$ value that makes $x^k = e$
so examples when looking at $D_4$ where $R = 90^\circ rotation$ and $F = horizontal flip$ :
- $|R^2| = 2$ since two rotations bring you halfway and so doing this twice will preserve it's footprint
- $|RF| = 2$ since when performing this on a piece of paper with numbered corners i got back to my original starting point after performing RF twice
but when moving into the proof of orders that are dealing with numbered groups i have trouble wrapping my head around what's going on..
Proposed with an exercise of examining $S_4$, when computing $|(12)(13)|$ or $|(1243)|$ i lack the intuition of seeing how this gets mapped and how we deduce the order of it.. (although my guess would be $|(12)(13)| = 4$ since it can be rewritten as $|(123)|$)
please let me know what you think; any and all help, input, and assistance is greatly appreciated!
You started with a slight error. There's a difference between the order of a group, and the order of an element. To illustrate, a favorite theorem of mine is Cauchy's theorem. It states that if $p$ is a prime dividing the order of the group, then the group has an element of order $p$.
But getting back to your question, actually $|(12)(13)|=|(123)|=3$, because a $3$-cycle has order $3$. More generally, an $n$-cycle has order $n$. Your intuition will develop as you move along.
Here's a useful fact: if $a,b\in S_n$ are disjoint, then $|ab|=\operatorname{lcm}(|a|,|b|)$. So for instance, let's look at $(12)(34)$. Since the transpositions $(12)$ and $(34)$ are disjoint, the order of their product is $2$. To contrast, in your example $(12)(13)$, the transpositions $(12)$ and $(13)$ are not disjoint, as they both "move" $1$.
That's all for now.