Finding the importance behind the proof that ${\rm ord}(bab^{-1})={\rm ord}(a)$.

Question

I recently completed the proof outlined here:

Trouble finishing up the proof that ${\rm ord}(bab^{-1})={\rm ord}(a)$.

and am trying to flesh out the meaning behind the proof. My basic understanding is as follows:

Firstly, I would think that by the property of closure of a Group, $\forall b \in G$, the composition $bab^{-1}$ will spit out every element of G as I cycle through different elements to serve as my "$b$" and "$b^{-1}$". This would then entail that $\forall x \in G{ \rm \ ord}(x) ={\rm ord}(a)$.

Now, clearly this cannot always be true (an easy counter example is $\mathbb Z_4$ where ${\rm ord} (1) =4$ but ${\rm ord} (2) =2$ )

So I went back into the assumptions that I made when deriving the ${\rm ord}(bab^{-1}) = {\rm ord}(a)$ proof...but cannot find any obvious assumption that alters my interpretation.

Therefore, the only issue that I can find with my interpretation is that the $bab^{-1}$ composition does not necessarily map to all elements of G as I cycle through different $b$'s ... e.g. ($cac^{-1}, dad^{-1},$ etc)...i.e. there must be some elements where $x\neq y$, but $xax^{-1} = yay^{-1}$

With this in mind, I am going to work forward from the following assumption and see what properties must emerge:

$x\neq y$, but $xax^{-1} = yay^{-1}$

Pre-multiplying by $x^{-1}$ and post-multiplying by $x$, we get:

$a=x^{-1}yay^{-1}x$

We could have $x^{-1}y=e$ and $y^{-1}x=e$, but this would be a contradiction because it implies that $x=y$

Therefore, the only way for $ a=x^{-1}yay^{-1}x$ is if $a$ is commutative with respect to $x$ and $y$.

With this in mind, it seems I can say the following: if no elements of $G$ are commutative, then $\forall x \in G {\rm \ ord}(x) ={\rm ord}(a)$.

Or more generally, if no elements in G are commutative, then all elements have the same order.

Is this correct?

Answer 1

That's incorrect, but let's explore what was valuable about your thinking.

Instead of considering conjugation by itself, consider the automorphism $\alpha_g: x \to g^{-1}xg$ (my class usually uses this instead of $gxg^{-1}$, but they're clearly equivalent). This is what's called an "inner automorphism", and it's identified with a particular element $g$.

Firstly, this immediately shows that $\operatorname{ord}(a) = \operatorname{ord}(g^{-1}ag)$, since any automorphism must map elements to other elements with the same order.

Secondly, you are correct when you say that in general, it's not true that it's possible to map every element to any other element with some inner automorphism. Using group action terminology, we'd say that "inner automorphisms do not act transitively on the set $G \setminus 1$.

Side note: In fact, there does not exist any nontrivial finite group $G$ so that its inner automorphisms act transitively on $G \setminus 1$. The proof of this is simple: In the rare case that the group of all automorphisms of $G$ act transitively on $G \setminus 1$, the group is a special type of abelian $p$-group: a direct sum of $C_p$ for some prime $p$. But then it's an abelian group, so the group of inner automorphisms is trivial and hence cannot act transitively. Try to prove this rigorously, it's a great exercise!

In general, all we know is the following statement: $\operatorname{Inn}G \cong G/Z(G)$. In English, the group of inner automorphisms is isomorphic to the group quotiented by its center. Try to prove this using the logic that you showed above!

A corollary of this fact is that $x = yz, z \in Z(G)$, if and only if $\alpha_x = \alpha_y$. In English, two elements produce the same inner automorphism if and only if they are in the same coset of $Z(G)$. So, if the center of $Z(G)$ is trivial (a.k.a none of the elements except $1$ are commutative), all the inner automorphisms are different.

However, this does not imply that the inner automorphisms act transitively on the set $G \setminus 1$. This is the flaw in your reasoning.

Hence it is possible for the elements to still have different orders. Look at $S_5$, for example, which has a trivial center (all nontrivial elements are not commutative), but plenty of elements have the same order, and plenty of elements have different orders.

Hopefully this was a helpful exploration, and a good demonstration of the power of thinking about conjugation in terms of inner automorphisms.