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I have

$$\sigma = \binom{1\ 2\ 3\ 4\ 5}{2\ 5\ 4\ 3\ 1} \in S_5$$

How do I show that $(1, 4)\sigma (1, 4)^{−1}$ is not an element of $\langle \sigma \rangle$?

(I don't know even what $(1, 4)\sigma (1, 4)^{−1}$ means so I need help with the basics)

N. F. Taussig
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user1162295
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1 Answers1

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$\sigma$ is a permutation which was defined in the problem statement. $(1,4)$ is the permutation which swaps $1$ and $4$ while leaving the rest alone. $(1,4)^{-1}$ is the inverse of that permutation (which feels a bit silly to write this way since $(1,4)^{-1}$ is just equal to $(1,4)$ until you realize we're trying to generalize). So, $(1,4)\sigma(1,4)^{-1}$ is the product (or composition) of the three permutations $(1,4),\sigma,$ and $(1,4)^{-1}$. Finally $\langle \sigma\rangle$ is the subgroup of $S_5$ generated by $\sigma$.

So, the question is whether $(1,4)\sigma(1,4)^{-1}$ is equal to one of the permutations in the subgroup generated by $\sigma$

This is trying to lead you to discovering whether the subgroup generated by $\sigma$ is normal or not. Perhaps you haven't studied normal subgroups yet, and this question was just there to plant the seed so that in a few more lessons you can look back and realize why this question was asked the way it was.

Suggested approach. What can you say about every element of $\langle \sigma\rangle$ with respect to the possible values that the number $3$ gets mapped to?

$3$ is always only ever going to be mapped to $3$ or $4$.

After evaluating $(1,4)\sigma(1,4)^{-1}$, what value does $3$ get mapped to here?

$3$ gets mapped to $1$ in the result of $(1,4)\sigma(1,4)^{-1}$

What can you conclude from that?

$(1,4)\sigma(1,4)^{-1}$ is not an element of $\langle\sigma\rangle$ and therefore $\langle\sigma\rangle$ is not a normal subgroup of $S_5$.

JMoravitz
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  • How do I find (1,4)(1,4)^-1? – user1162295 Mar 21 '23 at 13:38
  • @user1162295 It is unclear whether you are being taught to compose from right-to-left or from left-to-right. Both are common conventions. Thankfully, it doesn't matter here, they both result in the same. I recommend thinking of it by way of "element-chasing". Ask the question "How does the 'first' permutation affect the number $1$? How does the second permutation affect the result? How does the third affect the result of that*?" Here, the first maps $1$ to $4$. The second maps $4$ to $3$, and the third maps $3$ to $3$. Repeat that process for all numbers $1,2,3,4,5$. – JMoravitz Mar 21 '23 at 13:46
  • is (2 4 5)(1 3) correct? First I taped (1,4) to and then to (1,4)^-1 – user1162295 Mar 21 '23 at 14:02
  • $(2~5~4)(1~3)$, two maps to five here, not to four. – JMoravitz Mar 21 '23 at 14:04
  • hmm then I have done something wrong, is (1,4)= (425)(31) and (1,4)^-1 = (452)(31)? – user1162295 Mar 21 '23 at 14:16
  • What you just wrote is wrong and is surely missing a great deal... $(1,4)$ is equal to $(1,4)$... or written another way $\begin{pmatrix}1&2&3&4&5\4&2&3&1&5\end{pmatrix}$, or yet another way as $(1,4)(2)(3)(5)$... certainly not as $(4,2,5)(3,1)$. You seem to be trying to write $(1,4)\sigma(1,4)^{-1}$ and might be confusing $(1,4)\sigma(1,4)^{-1}$ with $((1,4)\sigma(1,4)^{-1})^{-1}$ – JMoravitz Mar 21 '23 at 14:21
  • In any event, when writing permutations in standardized form as the product of disjoint cycles, it is customary to arrange it such that the smallest number is in the front of each cycle. Rather than writing $(4,2,5)$ it is customary to write that as $(2,5,4)$ and so on... – JMoravitz Mar 21 '23 at 14:23
  • okey now I understand what you meant by saying switching 1 and 4 in the orgininal answer. It is that 1 -> 4 and 4->1 and the rest map to them self? not that 1 and 4 switch places in ? – user1162295 Mar 21 '23 at 14:37
  • when I map (1,4)(2)(3)(5) to (125)(34) and then back to (1,4)(2)(3)(5) I only get the same ((1,4)(2)(3)(5)) not (254)(13) – user1162295 Mar 21 '23 at 14:40
  • I will word this as though we are going right to left. Adjust if necessary to fit. We have three functions... $f,g$ and $h$ where we have $f(x)=h(x)=\begin{cases}4&\text{if }x=1\1&\text{if }x=4\ x&\text{otherwise}\end{cases}$ and $g(x)=\begin{cases}2&\text{if }x=1\ 5&\text{if }x=2\4&\text{if }x=3\3&\text{if }x=4\1&\text{if }x=5\end{cases}$. We are talking about $(f\circ g\circ h)$. We have that $(f\circ g\circ h)(x)=f(g(h(x)))$ and that $f(g(h(3)))$ for instance is equal to $f(g(3))$ which further simplifies to $f(4)$ which is equal to $1$... so $(f\circ g\circ h)(3)=1$ here – JMoravitz Mar 21 '23 at 14:44
  • Similarly, you can find all of the other values and can represent the resulting function $(f\circ g\circ h)$ as a permutation using your favorite way of writing it... whether in function form, two-line permutation form, or as a product of disjoint cycles... which if you did as a product of disjoint cycles you would have $(f\circ g\circ h)$ can be written as $(2~5~4)(1~3)$ or similar. – JMoravitz Mar 21 '23 at 14:46