Isomorphism preserves conjugacy class Let $f$ be an isomorphism such that $f: G\to G'$. For each $a\in G$, I need to show that $f (\text {cl}_G(a))=cl_{G'}(f(a))$ for $a$.
I write the definition of conjugacy:
$cl_G(a)=\{b \mid b=cac^{-1}\}, c\in G$, 
Now, 
$f(cl_G(a))= \big\{f(b) \mid f(b)=f(c)f(a)f(c^{-1})\big\} $,
How do I proceed from here? 
Any hints would be appreciated!
Thanks in advance!
 A: I really dislike the notation here. 
But if we've established that $f(\operatorname{cl}_G(a)) = \{f(b): f(b) = f(c)f(a)f(c^{-1})\}$, it should be a small leap to use the properties of homomorphisms to move the inverse outside of $f$, and write
$$f(\operatorname{cl}_G(a)) = \{f(b): f(b) = f(c)f(a)f(c)^{-1}\}.$$
How does this compare to $$\operatorname{cl}_{G'}(f(a)) = \{x: x = yf(a)y^{-1}\},$$
using the definition of $\operatorname{cl}_{G'}(f(a))$, where $x,y \in G'$? Because remember: in the original definition of the conjugacy class of $a$, the $b$ and $c$ were just dummy variables. Here $x$ and $y$ are the equivalent dummy variables in $G'$.
Well, since $x, y \in G'$, with $f: G \to G'$ a bijection (among other things), we can write $x = f(b)$ and $y = f(c)$ without loss of generality, because $x, y, b,$ and $c$ are all just dummy variables.

Again, I would strongly encourage you to get used to thinking of a conjugacy class as, for example, $\operatorname{cl}_G(a) = \{gag^{-1} : g \in G\}$ the set of things you get as you conjugate $a$ by all the elements in your group; as (our dummy variable) $g$ "runs over" $G$. It's just really bad form for the book to use an unqualified variable in a formula like that, that makes things confusing. It's also confusing to give the result of conjugation (originally $b$) a name when we never really need to refer to that name, it's just "some conjugate" of $a$. The lack of qualifiers/quantifiers is just really confusing in the original.
In "the standard" version of the proof, it'll turn out that as $c$ "runs over" $G$, it'll be obvious that $f(c)$ "runs over" $G'$:
$$f(\operatorname{cl}_G(a)) = \{f(c)f(a)f(c)^{-1} : c \in G\} = \{f(c)f(a)f(c)^{-1} : f(c) \in G'\} = \operatorname{cl}_{G'}(f(a)).$$
