2
$\begingroup$

Let $G= \langle \mu \rangle$ be the subgroup $S_4$ generated by $$\mu =\begin{pmatrix} 1 & 2 & 3 & 4\\ 2 & 3 & 4 & 1 \end{pmatrix}.$$ List all elements of $G$.

I believe this is just a 90 degree rotation counter clockwise (if we had a square of 1 in top left corner, 2 top right, 3 at bottom right and 4 at the bottom left). Or that is at least how I am thinking about it. If I follow that logic, then $G = \{\mu^0, \mu^1, \mu^2, \mu^3\} = \{\begin{pmatrix} 1 & 2 & 3 & 4 \end{pmatrix}, \begin{pmatrix} 2 & 3 & 4 & 1 \end{pmatrix}, \begin{pmatrix} 3 & 4 & 1 & 2 \end{pmatrix}, \begin{pmatrix} 4 & 1 & 2 & 3 \end{pmatrix}\}$

Is this the correct way of thinking about this? What are other ways of thinking about this problem?

$\endgroup$
2
  • 2
    $\begingroup$ Your answer is correct. I usually find cycle notation helpful, in which case you'd write $\mu = (1~2~3~4)$, meaning $\mu$ sends $1$ to $2, 2$ to $3, 3$ to $4$, and $4$ back to $1$. You then would write $\mu^2=(1~3)(2~4)$. $\endgroup$ Nov 6, 2021 at 3:52
  • 1
    $\begingroup$ Hmm thank you. Yeah that notation confuses me and I think is the reason why I have come up with different ways of thinking about this. I will review that right now. Again thank you. $\endgroup$ Nov 6, 2021 at 4:02

1 Answer 1

1
$\begingroup$

Your thought process is actually incorrect.

The symmetric group $S_4$ is a group consisting of functions, whose operation is function composition. The elements you are listing for G are not functions and are not elements of $S_4$. You need to write the elements in either the 2 row notation you started with or disjoint cycle notation as mentioned by Robert Shore to clearly communicate how you are defining the behavior of each function/element.

$\endgroup$
6
  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Nov 6, 2021 at 7:08
  • $\begingroup$ Oh wow, thank you this is helpful. There are some things about abstract algebra I am not getting. This helps clarify many things. Should I be thinking of almost everything as a function? $\endgroup$ Nov 6, 2021 at 20:18
  • $\begingroup$ No. Rather, the definition of each different kind of group should inform how you think of the elements in the group. The symmetric group $S_n$ has the set of all possible permutations with n letters/symbols as elements. Permutation is defined as a function that is both one-to-one and onto, whose co-domain and domain are the same[aka. the letters/symbols/numbers are mapped to the exact same letters/symbols/numbers]. $\endgroup$
    – HMR
    Nov 6, 2021 at 21:41
  • $\begingroup$ Consider the group $Z_8$. It's elements are defined as the numbers {0,1,2,3,4,5,6,7} with the operation addition modulo 8. The set {0,1,2,3,4,5... elements are not functions but numbers, because the group defined them that way. Look at the definition of whatever group you are playing around with to know how you should consider/label/display the elements. Hope this helps. :) $\endgroup$
    – HMR
    Nov 6, 2021 at 21:48
  • $\begingroup$ Oh okay. So then $S_4$ are all permutations functions of set size 4? I thought it was just all permutations of $1,2,3,4$, but they are actually permutation functions. $\endgroup$ Nov 6, 2021 at 22:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .