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I need help showing this:

Let G be a finite group such that for every $x$, $y$ in G, $x\neq 1$ and $y\neq 1$, 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?

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I'll note that the condition that $G$ is finite is essential. There are infinite groups in which any two non-identity elements are conjugate (they can be constructed as a sequence of HNN extensions). – Arturo Magidin Apr 28 '12 at 3:44
up vote 12 down vote accepted

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.

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To check your understanding: what's $\mathrm{stab}_G(x)$ in this case? – Jonathan Apr 28 '12 at 1:00
$stab_G(x)$ is the set of all elements of $G$ that $gx=x$, that is, $stab_G(x)=\{g\in G | gx=x\}$. Right? – Marra Apr 28 '12 at 1:15
I know now why I couldn't have a clue on this before: $x$ and $y$ are conjugate if there is an element $z$ other than $1$ such that $x=zyz^{-1}$. I forgot to exclude the 1! – Marra Apr 28 '12 at 1:17
@GustavoMarra That's the right definition, yes (provided one interprets your notation correctly). $\mathrm{stab}_G(x)$ is the set of all elements of $G$ such that $g \cdot x = x$ where "$\cdot$" denotes the action. So in the case of the conjugation action, this is equivalent to saying $\mathrm{stab}_G(x) = \{ g \in G \ | \ gxg^{-1} = x \}$ where (for lack of a better term) "placement of letters next to each other" corresponds to the group multiplication law. What I mean to ask is "can you list the elements in the stabiliser of $x$ for a particular $x$? (for this particular action)" – Jonathan Apr 28 '12 at 1:18

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 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 elements of the center. Thus $|G|=2$.

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This is fine, but it uses Sylow's theorems... and I'm supposed to solve that without them. – Marra Apr 28 '12 at 1:28
Yeah, Jonathan's solution is nicer anyhow, but I figured it would be useful to see a proof that makes good use of the some of the bigger theorems you find in a first course on group theory. – Brett Frankel Apr 28 '12 at 1:32

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