# Proving Equivalence Relation on a set of functions

Consider the set of functions from some set X $$\rightarrow$$ X. The relation fRg is defined by $$g = a^{-1} \circ f \circ a$$ for some function $$a$$ that is bijective from X $$\rightarrow$$ X. I am wondering how to prove the equivalence relation, and to find the equivalence class for f(x) = x.

Reflexive: Look at $$f(x) = a^{-1}(f(a(x)))$$. Then take the bijective function $$a(x) = x$$, and clearly $$f(x) = f(x)$$ for all $$x \in X$$. So the relation is reflexive.

I'm having trouble proving the others. For example, with symmetry, I am unable to find a way to get from $$g(x) = a^{-1}(f(a(x))) \Longrightarrow f(x) = a^{-1}(g(a(x)))$$.

• If $g = a^{-1}\circ f\circ a$, then $f = a\circ g\circ a^{-1}$. Can you prove transitivity? – amsmath Oct 6 '19 at 1:50
• My current efforts with transitivity have left me stuck; assume two relations fRg and gRh. Then I can write $h = a^{-1} \circ g \circ a$. Then replace the $g$ in this equation to get $h = a^{-1} \circ a^{-1} \circ f \circ a \circ a$, which does not give the result I hoped for since the function $a$ is there two times. I feel like I'm misunderstanding something important – sjrr00 Oct 6 '19 at 2:10
• Nope. The '$a$' connecting $h$ and $g$ might be another function than $a$, so you should give it another name, say $b$. Note furthermore that $(a\circ b)^{-1} = b^{-1}\circ a^{-1}$. – amsmath Oct 6 '19 at 2:11
• oh gosh, thanks for pointing that out. Using the property that you told me, I was able to get $(a \circ b)^{-1} \circ f \circ ( a \circ b)$ which corresponds to the what the relation needs. – sjrr00 Oct 6 '19 at 2:33
• If you put $b=a^{-1}$, then $f = b^{-1}\circ g\circ b$, as desired. ;-) – amsmath Oct 6 '19 at 2:42

Reflexive:$$f=id f id^{-1} \implies fRf$$
Symmetric :$$f=a^{-1}ga \implies g=afa^{-1}$$
Transitive: $$f=a^{-1}ga$$, and $$g=b^{-1}hb$$ then $$f=a^{-1}b^{-1}hba = (ba)^{-1}h (ba)$$