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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.

1 Answers 1