Htykuut

I need help showing this:

Let G be a finite group such that for every $x$, $y$ in G, $xneq 1$ and $yneq 1$, we have that $x$ and $y$ are conjugates. Under those conditions, G must have order 1 or 2.

This is under the topic "actions of groups on sets", but I couldn't figure out a way to start it. Since every element is conjugate, then G must have only one conjugation class, which is itself, but how can this information help?

If $G$ is the trivial group, then the condition holds. So suppose $G$ isn't the trivial group.

$G$ acts on itself by conjugation, i.e. we have an action $G times G to G$ where $(g,h) mapsto ghg^{-1}$. Take an element $x in G$. Since any two non-identity elements are conjugate, $mathrm{orb}_G(x) = G backslash {e}$, so $|mathrm{orb}_G(x)| = |G| - 1$. But the orbit-stabiliser theorem tells us that the size of an orbit divides the order of the group. So $|G|-1 $ divides $|G|$. So $|G|$ must be 2.

Here's another proof, not quite as elegant, but with a different flavor that I feel is also worth seeing. In this proof I use a few of the theorems you may have seen in the "group actions" section of the book:

Since conjugate elements have the same order, all nonidentity elements have the same order. Thus only one prime number, $p$, divides the order of the group, since for every prime dividing $|G|$ we have an element whose order is that prime (This follows from Cauchy's Theorem and the Sylow Theorems, which I expect you'll encounter soon). But we know $p$-groups have nontrivial centers (by the class equation, one of the most important elementary results using group actions), and elements of the center are their own conjugacy classes. Since the order of the center is at least $p$ and the center has at most one nonidentity element, $p=2$, and $|Z(G)|=2$. If $|G|$ is 4 or more, then we have elements not in the center, which cannot be conjugate to elements of the center. Thus $|G|=2$.