Abstract Algebra - Finite Group Let G be a non-trivial finite group. For every $a,b \in G$ that are not identities, there exist $c \in G$ such that $b=c^{-1}ac$. Show that $|G|=2$.
 A: We can answer this question as follows:
Define $a \equiv b$ if $\exists c $ such that $b=c^{-1}ac$. Note that this is an equivalence relation on the group $G$, and hence divides it into equivalence classes of the form $[a]=\{ c^{-1}ac\  | c \in G\}$. We are given that every $a$ and $b$ that are not identities  are such  that $a \equiv b$. Thus, the equivalence class of $a$ consists of all elements except the identity, so it's order is $|G|-1$. 
Now the class equation comes to our rescue. What it basically says is that the size of the equivalence class divides the order of the group! So $|G|-1 $ divides $|G|$, and so $|G|=2$, since this is the only finite quantity with this property!
Please comment back if  you don't know the class equation: I can edit my answer later on!
A: It suffices to show that $G$ is abelian. Suppose not. Since every nonidentity element is conjugate to every other, every nonidentity element has the same order. By Cauchy's theorem this implies that $G$ must be of prime power order because otherwise there would be two elements whose orders are distinct primes.
Note that since $G$ is not abelian, there exists a nontrivial commutator element. It follows by the conjugacy condition and normality of the commutator subgroup that $G$ is perfect. But groups of prime power order are solvable, which gives us a contradiction. Thus $G$ is abelian, and since every nonidentity element is conjugate it must have order 2 if it is not trivial.
