Q:

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!

