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In Exercises 40 through 43, let $A$ be a set, $B$ a subset of $A$, and let $b$ be one particular element of $B$. Determine whether the given set is sure to be a subgroup of $S_A$ under the induced operation. Here $\sigma[B] = \{ \sigma(x) \mid x \in B \}$.

40. $\{ \sigma \in S_A \mid \sigma(b) = b \}$
41. $\{ \sigma \in S_A \mid \sigma(b) \in B \}$
42. $\{ \sigma \in S_A \mid \sigma[B] \subseteq B \}$
43. $\{ \sigma \in S_A \mid \sigma[B] = B \}$


p 85 #41 Solution   This is not necessarily a subgroup of $S_A$. Consider the following example: let $A = \{ 1,2,3 \}$, $B = \{ 1,2 \}$, $b = 1$ and $\sigma = \begin{pmatrix}1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}$ Then $\sigma (1) = 2 \in B$. However $\sigma^{-1} ( 1 ) = 3 \notin B$ which shoes that $\{ \sigma \in S_A : \sigma(b) \in B \}$ is not closed under the operation of taking inverses. $\Box$

p 85 #42 Solution   This is not necessarily a subgroup of $S_A$. Consider the following example: Let $A = \mathbb{Z}$, $B = \mathbb{N} = \{ n \in \mathbb{Z} \mid n > 0 \}$, and $\sigma$ defined by $\sigma (n) = n+1$. Clearly, $\sigma$ is one-to-one and onto so $\sigma \in S_A$. Also, if $n > 0$, then $\sigma (n) > 1$ and so $\sigma [ \mathbb{N} ] \subseteq \mathbb{N}$. However, $\sigma^{-1} ( n ) = n-1$ and so $\sigma^{-1} [ \mathbb{N} ] \not\subset \mathbb{N}$ since $0 \in \sigma^{-1} [ \mathbb{N} ]$. $\Box$

  1. How do you magically envisage and envision 41 and 42 will not be subgroups?

    I thought about this but I'm unsettled. By the One-step Subgroup Test, $\{\sigma \in S_A : predicate \}\text{ is a subgroup of }S_A \iff \sigma_1\sigma_2^{-1} \in \{\sigma \in S_A : predicate \} $ for all $\sigma_1,\sigma_2 \in S_A$.

    For 41, hence take $\sigma_1, \sigma_2 \in S_A$ such that $\sigma_1(b), \sigma_2(b) \in B$. Need $\sigma_2^{-1}(b) \in B$. $B$ is only a subset and not a group, hence $\sigma(b) \in B \not\implies \sigma^{-1}(b) \in B $. Hence envisage this false?

    For 42, hence take $\sigma_1, \sigma_2 \in S_A$ such that $\sigma_1(B), \sigma_2(B) \subseteq B$. Need $\sigma_2^{-1}(b) \subseteq B$. $B$ is only a subset and not a group, hence $\sigma(b) \in B \not\implies \sigma^{-1}(b)\subseteq B $. Hence envisage this false?

  2. How are 41 and 43 related? I can prove 43 is a subgroup. However I am confounded why 41 is false and 43 is true.

  3. For 41, here's Fraleigh's solution:

    No, the set need not be closed under the operation if B has more than one element.
    Suppose that $σ$ and $μ$ are in the given set, $\quad$ $b,σ(b) ∈ B$, $\quad$ but $\color{brown}{μ(σ(b)) \notin B}.$
    Then $(μσ)(b) = \color{brown}{μ(σ(b)) \notin B}$, so $μσ$ is not in the given set.

    How do you envisage and envision this counterexample? It looks more magical than the first solution.

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  • $\begingroup$ Please typeset your question using $\TeX$. Screenshots can be diffcult to read and aren't searchable. $\endgroup$ – Ayman Hourieh Jan 11 '14 at 19:05
  • $\begingroup$ Voting to reopen. I'm not sure that I'm best equipped to answer this, but I think that there is an answerable question in there somewhere. The point of this exercise is IMVHO to build a bit of intuition about how permutation groups move objects of sets. Sometimes there are relatively subtle things at play. As a final exercise ponder the following: If in #42 we add the assumption that the subset $B$ is finite, then the conclusion changes in the sense that we again get a subgroup! Anyway, your plight is shared by many beginning students of permutation groups, so the answers should be pedagogical $\endgroup$ – Jyrki Lahtonen Jan 19 '14 at 8:50

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