This is another homework question I can't figure out.

For $|G|$ even, $\forall x\in G\exists b\in G\setminus{\{x^{-1}\}}$ such that $bxb = x^{-1}$

I tried to toy with associativity but to no avail. Also, I can't see the relevance of $|G|$ being even. Any hint (I'm not here for the solution) is appreciated.

Thanks for your attention.

Update: I thought I would show you the proof I've written since that's the least I can trade for the effort you took to help me. I'm eager for any kind of feedback so, if you're about to comment on it, show no mercy. I'm trying to improve.

Let $x \in G$. There's left to prove that there exists $b \in G\setminus{\{x^{-1}\}}$ such that:
$bxb = x^{-1}$
$\Longleftrightarrow (bxb)x = x(bxb) = e$
$\Longleftrightarrow (bx)bx = xb(xb) = e$
$\Longleftrightarrow (bx)^{2} = (xb)^{2} = e$
$\Longleftrightarrow bx = (bx)^{-1}$ and $xb = (xb)^{-1}$
$\Longleftrightarrow bx = x^{-1}b^{-1}$ and $xb = b^{-1}x^{-1}$

Let $a \in G\setminus{\{e\}}$ such that $a = a^{-1}$. (We know such $a$ exists by a previous result). Let $b = ax^{-1}$. Then:
$bx = x^{-1}b^{-1}$ and $xb = b^{-1}x^{-1}$
$\Longleftrightarrow ax^{-1}x = x^{-1}(ax^{-1})^{-1}$ and $xax^{-1} = (ax^{-1})^{-1}x^{-1}$
$\Longleftrightarrow a = x^{-1}xa^{-1}$ and $xax^{-1} = xa^{-1}x^{-1}$
$\Longleftrightarrow a = a^{-1}$ and $xax^{-1} = xax^{-1}$

What do you think of it?

  • 1
    $\begingroup$ This is equivalent to showing $(bx)^2=e$ ($e$ being the identity), which is only possible in a group of even order. This is because the order of an element has to divide the order of the group. $\endgroup$ – Cheerful Parsnip Feb 17 '11 at 1:01
  • $\begingroup$ @Jim: the order of an element has to divide the order of the group, but that doesn't imply that there are elements of any order that divides the order of $G$ (well, in this case it does, but I think this is an early exercise, that doesn't require use of Cauchy's theorem). $\endgroup$ – Weltschmerz Feb 17 '11 at 1:07
  • $\begingroup$ @Weltschmerz: I agree. I said it's only possible in a group of even order, not that it always happens in a group of even order. (Although it does.) :) $\endgroup$ – Cheerful Parsnip Feb 17 '11 at 2:11
  • $\begingroup$ @Maria: The proof is generally okay, with two problems: (i) in the first part, your first equivalence is not quite right, since you are not taking into account the condition $b\neq x^{-1}$; you need to add a clause (and carry it around until the last equivalence) that says that $bx\neq e$ (or that $b\neq x^{-1}$). (ii) Once you have that, the second part can be shortened substantially: once you set $b=ax^{-1}$, all you have to do is note that $bx = a = a^{-1}=(bx)^{-1}$, and that $bx\neq e$, so by part (i), you are done. You don't have to repeat the entire chain of equivalences with the values $\endgroup$ – Arturo Magidin Feb 17 '11 at 16:01
  • $\begingroup$ +1 for adding your answer after you figured it out, to receive comments. $\endgroup$ – Arturo Magidin Feb 17 '11 at 16:03

Here are some hints, read them one at a time if you want to keep looking yourself :-)

1 - Saying |G| is even means the group contains some $a$ that satisfies $a^2 = 1$ but $a\neq 1$.

2 - For given $x$ you should try to find $b$ such that $(bx)^2 = 1$

3 - So you could attempt find $b$ such that $bx = a$

4 - This means $b = ax^{-1}$.


Say $|G|$ is even. There exists $a$ such that $a^2 =1$, $a\neq 1$. (See arguments in the comments.) Now take any $x\in G$ and let $b := ax^{-1}$. Indeed, this $b$ satisfies

  • $b\neq x^{-1}$, because $b = x^{-1}$ would yield the contradiction $bx = a = 1$
  • $ bxb = ax^{-1}xax^{-1} = aax = x $ as required.
  • $\begingroup$ A stronger form of #1 is that the number of elements $a$ such that $a^2=1$ is even. $\endgroup$ – lhf Feb 17 '11 at 1:34
  • $\begingroup$ Are all these tools available in that exercise? And can we simply use Sylow's theorem instead of Cauchy's theorem?Thanks for any clarification. $\endgroup$ – awllower Feb 17 '11 at 4:12
  • $\begingroup$ @awllower: Sylow's Theorems are more advanced than Cauchy's theorem; in fact, you usually need Cauchy's theorem to get started on Sylow's. But you don't need either in this case: there is an elementary counting argument to show that if $|G|$ is even, then there must be elements of order $2$: just define an equivalence relation on $G$ by $x\sim y$ if and only if $x=y$ or $x=y^{-1}$, note that every equivalence class contains either $1$ element (if the element is the identity or of order $2$) or $2$, and conclude that the number of equivalence classes with 1 element is even. $\endgroup$ – Arturo Magidin Feb 17 '11 at 5:13
  • $\begingroup$ Marvelous, I think I now understand the whole thing, thanks, @Arturo Magidin. $\endgroup$ – awllower Feb 17 '11 at 8:40
  • $\begingroup$ @Myself: I followed one by one and eventually, when I read hint 4, I got it! I should have suspected the $a^{2} = e$ result which we had to show in a previous exercise (and Andrea Mori kindly anticipated that in math.stackexchange.com/questions/20066/…). Thanks for taking the effort to give hints this way! $\endgroup$ – Marla Feb 17 '11 at 14:04

Hint. $bxb = x^{-1}\Longleftrightarrow bx = x^{-1}b^{-1}\Longleftrightarrow bx = (bx)^{-1}\Longleftrightarrow bx$ has order $1$ or $2$.

Excluding $b=x^{-1}$ guarantees that $bx\neq e$.

  • $\begingroup$ Your's very succinct hint really gave me insight! I am even surprised I understood it immediately! Thanks for your remark! $\endgroup$ – Marla Feb 17 '11 at 14:07

What the claim tells you is that for any element $x$ of the group $G$, there is some $b$ in $G$ (other than $x^{-1}$) such that $bx$ is its own inverse. Try to see what happens if you "try" all $b$'s (other than $x^{-1}$) in $G$ and never obtain, in the product $bx$, an element which is its own inverse. (Also note that when you run through all possible $b$'s in the product $bx$, you're actually covering the whole group -- this is because of the cancellation law). Think about the relationship between some elements being their own inverses (or, equivalently, having order 2) and the order of the group being even.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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