Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is an assignment question I received a week ago.

A function $f:\{1, 2, \dots ,n\} \to \{1, 2, \dots, n\}$ which is a bijection is also called a permutation. Let $P_n$ be the set of all permutations on $\{1, 2, \dots , n\}$. Define the relation $\sim$ on $P_n$ by $f\sim g$ iff there exists a permutation $h$ such that $f = h \circ g \circ h^{-1}$.

a) Need to show that it is an equivalence relation.

My Approach:

For a relation to be an equivalence relation, it must be reflexive, symmetric, and transitive.

Reflexive:

Let $f \sim f \Longleftrightarrow$ there exists $h$ such that $f = h \circ f \circ h^{-1}$.

My question (this may be silly) but is $h \circ f \circ h^{-1} = f$? I claimed that it was, and thus if $h \circ f \circ h^{-1} = f$, then it is reflexive.

Symmetric:

Let $f \sim g \Longleftrightarrow$ there exists $h$ such that $f = h \circ g \circ h^{-1}$. Let $g \sim f \Longleftrightarrow$ there exists $h_2$ (may be a different $h$) such that $g = h \circ f \circ h^{-1}$.

If $f = h \circ g \circ h^{-1}$ then $f = g$.

And since $g = h \circ f \circ h^{-1}$ then $g = f$.

Clearly this is reflexive.

I would like to know if I am in the right direction.

Regards,

Julian.

share|improve this question
add comment

1 Answer 1

You are generally going in the right direction. For reflexive, the permutations do not commute, so $h \circ f \circ h^{-1}$ does not necessarily equal $f$. But can find a specific $h$ that works. For symmetric, when you write $g$ you should write $g = h_2 \circ f \circ h_2^{-1}$. Again, because the permutations do not commute you cannot conclude (and it is not generally true) that $f=g$. But given $f = h \circ g \circ h^{-1}$ you should be able to find an $h_2$. For transitive, you assume $f = h \circ g \circ h^{-1}$ and $g= h_2 \circ k \circ h_2^{-1}$ Now can you find $h_3$ such that $f = h_3 \circ k \circ h_3^{-1}$?

share|improve this answer
    
Thank you for the guidance, Ross. –  Julian Park Oct 31 '12 at 2:34
    
If h is the identity function, then would h∘f∘h−1 = f since h: A -> A would be the same as it's inverse? Would that be a specific h that works? Or would it be a different case? –  Julian Park Nov 2 '12 at 2:19
    
@JulianPark: That is exactly what I was thinking. You have shown $f \sim f$ as you wanted. Another that works is $f$ itself, as does $f^{-1}$ (which need not be distinct from $f$) –  Ross Millikan Nov 2 '12 at 3:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.