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Problem (The Number Of Conjugates Of A)

Let $G$ be a group and let $a\in G$. Prove that the number of conjugates of $a$ equals the index of the centralizer of $a$ in $G$. In symbols, we write this as $|\text{cl}(a)|=|G:C(a)|$.

Hint: Recall that the index of $H$ in $G$ is the number of distinct cosets of $H$ in $G$.

Build a map from $\text{cl}(a)$ to the collection of cosets of $C(a)$ in $G$. Try sending $x^{-1}ax\in \text{cl}(a)$ to the coset $C(a)x$. You'll have to show that this map is well defined, meaning that if $x^{-1}ax=y^{-1}ay$, then you must have $C(a)x=C(a)y$. You'll need to then show that this map is a bijection. I would suggest you look up the properties of cosets to see what is equivalent to saying $Hx=Hy$.

One of the easiest ways to show that two sets have the same number of elements is to produce a bijection from one the others.

On a side note, the map you create is NOT a homomorphism. You won't be able to prove that it is. Luckily, you don't need to.

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