0

In Pinter's "A Book of Abstract Algebra", Chapter 13 Exercise J1 asks the reader to prove that the relation $\sim$ on set $A$, where $\sim$ is defined as "$u \sim v\ $ iff $\ g(u)=v$ for some $g \in G$", is an equivalence relation.

To provide further context, $g \in G$ where $G$ is a subgroup of the symmetric group $S_A$. Further, $u,v \in A$.

I am comfortable with these styles of proofs (reflexivity, symmetry, transitivity), but would like to use this particular exercise to ask two questions about notation.

For the symmetry part of the proof, I need to prove the below implication:

If $\exists g \in G \backepsilon g(u)=v$, then $\exists h \in G \backepsilon h(v)=u$

This is straightforward and one realizes that $h$ is simply $g^{-1}$. However, I have a question regarding how to write the implication.

Specifically, does the following implication (where $h$ is replaced with $g$) denote the exact same thing?

If $\exists g \in G \backepsilon g(u)=v$, then $\exists g \in G \backepsilon g(v)=u$

Or does this force the $g$ in the antecedent to be the exact same $g$ that is in the consequent? (or does the value of $g$ from the antecedent "reset" when looking at the $g$ in the consequent)

With that question out of the way, I want to now ask how one would reframe the definition of $\sim$ such that one intentionally requires that the $g$ in the antecedent is the same as the $g$ in the consequent.

Said differently, what would I need to change in the definition of $\sim$ to make the "symmetry implication" read as follows:

If there is a specific element $g_1 \in G \backepsilon g_1(u)=v$, then for that same element $g_1$, $g_1(v)=u$

S.C.
  • 5,272
  • 2
    Use $\sim$ for $\sim$. – Shaun Dec 13 '19 at 19:07
  • 1
    They are equivalent. For checking, a proof about such orbits is here.. It's not clear what you mean in the final paragraph. Please elaborate. – Bill Dubuque Dec 13 '19 at 21:06
  • Hmmm...I made an edit that hopefully clarifies. I am basically confused as to when the $g$ that is used in the antecedent should be interpreted as the exact same $g$ that is used in the consequent. Is that clearer? Basically I'm asking for the distinction between the two first highlighted statements (which you said were equivalent) and the 3rd (last) highlighted statement. – S.C. Dec 13 '19 at 21:34
  • 1
    Comment (not an answer). I much prefer (both as an author and as an instructor ) proofs written out mostly in English rather than with formal logical statements. Doing that might help with your statements - think about whether they read right. – Ethan Bolker Dec 13 '19 at 21:40
  • If it is the same $g$ then we can deduce that $g^2 = 1$ by $,u = gv = ggu,,$ which isn't true in most groups. – Bill Dubuque Dec 13 '19 at 22:57
  • @BillDubuque that is sort of my point. My question is therefore how would I know that the implication should be written in the form of the 1st/2nd highlighted statement as opposed to the 3rd highlighted statement? (i.e. what part of the $\sim$'s definition is telling me to pursue the 1st/2nd highlighted statement instead of the 3rd highlighted statement) – S.C. Dec 14 '19 at 00:41
  • And, to reference my actual question in the post, how would I alter $\sim$'s definition to indicate that someone should pursue the 3rd highlighted statement instead of the 1st/2nd highlighted statement – S.C. Dec 14 '19 at 00:48
  • 2
    Because the scope of the existentially quantified symbol $g$ in the definition of the equivalence relation does not extend beyond the expression it encloses. Any other external use of the symbol $g$ is unrelated. Indeed we could change the name of $g$ in the definition and the definition remains the same. It's a "dummy variable" - its particular name doesn't matter. They are called "bound" (vs. "free") variables. Search on those terms to learn more. – Bill Dubuque Dec 14 '19 at 00:48
  • 1
    To get the same $,g,$ in both equations you put them both in the same quantifier scope, e.g. $,u\approx v!!\overset{\rm def!!}\iff! \exists g\in G\ [,gu = v\ &\ u = gv,]\ \ $ – Bill Dubuque Dec 14 '19 at 01:07
  • Ohhhh, interesting. That definition is intrinsically symmetric, correct?

    i.e. if $\exists g \in G : gu=v$ and $u=gv$, then $\exists g \in G : gv=u$ and $v=gu$ is inherently a true statement because it is effectively saying "If $A$ then $A$"

     – S.C. Dec 14 '19 at 02:26
  • 1
    Yes, if we call if $P(u,v)$ then $P(u,v)\iff P(v,u)$. Note that generally you need address users explicitly by writing @ if you wish them to receive inbox notifications. I saw your prior message only because I still had the page open. – Bill Dubuque Dec 14 '19 at 18:44

0 Answers0