Is there some flaw in this reasoning?

Question

Problem statement: if $G$ is a group and $a,b\in G$, prove that if $|ab|=k$, then $|ba|=k$.

I'm going to assume one thing: that $|a|$ means $|\langle a \rangle|$, that is, the size of the subgroup generated by $a$.

I argued that the size of each subgroup is less than or equal to the size of the other subgroup, and that therefore, their sizes are equal.

($|ba|\leq |ab|$) Every element $g\in \langle ba \rangle$ has the form $(ba)^n$, where $n\in \mathbb{Z}$. This element can be generated by $n+1$ compositions of $ab$: $(ab)^{n+1}= a(ba)^nb$. This means that every element in $\langle ba\rangle$ can be generated in $\langle ab\rangle$ and that therefore $\langle ab\rangle$ must have at least as many elements as $\langle ba\rangle$.

I used the same reasoning to conclude $|ab|\leq |ba|$.

What is wrong with this reasoning (if at all)?

Answer 1

Another way, perhaps shorter. First, prove the easy

Lemma: For any elements $\,x\,,\,g\,$ in a group $\,G\,$ , we have that $\,|x|=|x^g|=|g^{-1}xg|\,$

Well, we're then done since $\,ba=(ab)^a\,$

Answer 2

As Thomas says, it is false that the elements in $\langle ab \rangle$ are necessarily in $\langle ba \rangle$, but your argument can be recovered :

Consider the function $f : \langle ab \rangle \to \langle ba \rangle$ defined by $f(g) = bga$. It is well-defined because any element $g \in \langle ab \rangle$ can be written $(ab)^k$, and then we see that $b(ab)^ka = (ab)^{k+1}$ is in $\langle ba \rangle$, as you noted. Since we're in a group, $f$ is injective : if $f(g) = f(h)$, you can cancel $a$ and $b$ to obtain $g=h$. This shows that $| \langle ab \rangle| \ge |\langle ba \rangle|$.

Answer 3

As I said in comments, it is not true that every element in $\left<ba\right>$ is in $\left<ab\right>$.

Your argument can be made rigorous as follows.

Let $S = \left<ab\right>$ and $T=\left<ba\right>$. Then to prove that $|ab|=|ba|$ you need to find a $1-1$ and onto function between these two sets.

What you have shown is that if $g\in T$, then $agb\in S$.

So let's define that function $f:T\rightarrow S$ by $f(g)=agb$.

Then $f$ is $1-1$, because if $ag_1b = ag_2b$ you can show $g_1=g_2$ by multiplying by $a^{-1}$ on the left and $b^{-1}$ on the right.

Is $f$ onto? You can show that if $h\in S$, then $h=(ab)^k$ for some $k$ and hence that $h=a(ba)^{k-1}b = agb$ where $g=(ba)^{k-1}\in T$, so $h=f(g)$ for some $g\in T$, and $f$ is onto.

So $f$ is $1-1$ and onto, and hence $S$ and $T$ have the same size, so $|ab|=|ba|$

This theorem is usually proved in other ways, using the alternate definition I gave above for $|g|$ (the smallest positive $k$ such that $g^k=1$.) I wanted to present a proof that followed your argument, roughly.