5
$\begingroup$

Isn't it obvious for $|a|=|a^{-1}|$, since $\langle a\rangle = \langle a^{-1} \rangle$?

For $|ab|=|ba|$, I think we should go like this: $e=(ab)^n\Rightarrow e=(ab)(ab)^{n-1}\Rightarrow (ab)^{-1}=(ab)^{n-1}\Rightarrow b^{-1}a^{-1}=(ab)^{n-1}\Rightarrow a^{-1}=b(ab)^{n-1}$ But I get stuck here, since I don't know whether to get from here to $e=(ba)^n$.

For $|a|=|cac^{-1}|$, do I have to show that $\langle a\rangle = \langle cac^{-1}\rangle$?

$\endgroup$
2
$\begingroup$

For 1) Let $|a|=n$ , i.e. $a^n=e$

Now take inverse both side $(a^n)^{-1}=e^{-1}$ which gives $(a^{-1})^n=e$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

For the second one:

Notice that as $(ab)^n=1$. Then we have $(ab)^{n-1}=(ab)^{-1}=b^{-1}a^{-1}$.

Now consider $(ba)^n$: $$(ba)^n=b(ab)^{n-1}a=bb^{-1}a^{-1}a=e.$$

Hence $|ba| \Big\lvert |ab|$.

Now think, from the last computations, why one must have equality.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Oh, you posted your own answer while I was writing. $\endgroup$ – Diego Robayo Jul 11 '15 at 16:36
  • $\begingroup$ Thanks, but if you've written the complete answer, post it and I'll choose your answer as the accepted one rather than mine. :) $\endgroup$ – Mill Jul 11 '15 at 19:14
1
$\begingroup$

If I have understood your question correctly, and I am not misinterpreting your notation. The last part utilizes the second part. If for all $a_1,a_2 \in G$ we have $$ |a_1a_2| = |a_2a_1|$$ Then let $a \in G$ and $c \in G$ be arbitrary, and $a_1 = c$ and $a_2 = ac^{-1}$. Now use the second property displayed above.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Simple and smart, good one :) $\endgroup$ – Mill Jul 11 '15 at 16:28
1
$\begingroup$

All of the answers and comments helped me, but here is my own answer which brings a complete solution in one place.

First of all, we need to recall the following theorem.

Theorem 3.4.iii (XXIX.iii), Algebra (Hungerford)

Let $G$ be a group and $a\in G$. If $a$ has finite order $m>0$, then $m$ is the least positive integer that $a^m=e$.

Now $|a|=|a^{-1}|$, because

$|a|=n\Rightarrow a^n=e\Rightarrow (a^n)^{-1}=e^{-1}\Rightarrow (a^{-1})^n = e\Rightarrow |a^{-1}|=n\Rightarrow |a|=|a^{-1}|.$

And I also think I can say this statement is true because $\langle a\rangle = \langle a^{-1} \rangle$.

$|ab|=|ba|$ because

$(ab)^n=e\Rightarrow (ab)(ab)^n(ab)=(ab)(ab)=a(ba)b\Rightarrow (ab)^n(ab)^{n^2}(ab)^n=a^n(ba)^nb^n\Rightarrow e=a^n(ba)^nb^n\Rightarrow (ab)^{-n}=(ba)^n\Rightarrow ((ab)^n)^{-1}=(ba)^n\Rightarrow e=(ba)^n$

Hence, $|ab|=|ba|$.

And finally, $|a|=|cac^{-1}|$, because from the statement we proved right above, we know that $|xy|=|yx|, \forall x,y\in G$. Let $x=c$ and $y=ac^{-1}$, so we've got $|yx|=|ac^{-1}c|=|a|$ and $|xy|=|cac^{-1}|$, yielding $|a|=|cac^{-1}|$.

hacked :)

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.