Re: finite group of odd order has no nontrivial element conjugate to its inverse

On Thu, 06 Oct 2005 00:19:04 -0700, quasi <quasi@xxxxxxxx> wrote:

>On Wed, 05 Oct 2005 23:04:01 -0700, quasi <quasi@xxxxxxxx> wrote:
>
>>On Thu, 06 Oct 2005 02:31:07 GMT, Kira
>>Yamato<kirakun@xxxxxxxxxxxxxxxxxx> wrote:
>>
>>>
>>>quasi <quasi@xxxxxxxx> wrote:
>>>>
>>>>
>>>>On Thu, 06 Oct 2005 01:13:18 GMT, Kira
>>>>Yamato<kirakun@xxxxxxxxxxxxxxxxxx> wrote:
>>>>
>>>>>
>>>>>Need some help in proving that
>>>>> For G a finite group of odd order, then if there exists g and h in G
>>>>>such that h^{-1} g h = g^{-1},
>>>>> then
>>>>> g = e, the identity element.
>>>>>
>>>>>Thanks.
>>>>>-kira
>>>>
>>>>Hint:
>>>>
>>>>First show this:
>>>>
>>>>In any group G, if an element g has finite order, then for any h in G,
>>>>h^{-1} g h has the same order as g. In other words, an element has
>>>>the same order as any of its conjugates.
>>>>
>>>>In the context of the given problem, the above principle implies that
>>>>g and g^{-1} have the same order.
>>>>
>>>
>>>Err.. I thought in *any* group, for every element g in it, g and its
>>>inverse must have the same order anyway.
>>
>>Yes -- you're absolutely right -- I wasn't thinking carefully.
>>
>>So my hint has no power here -- sorry.
>>
>>quasi
>
>Well, it's probably simple, but I don't see it.
>
>As a plan for discovery, how about this ...
>
>Consider the subgroup <g,h> of G generated by g and h.
>If G has odd order, so does every subgroup, hence the subgroup <g,h>
>is a group of odd order, and the given relation between g and h still
>holds. Also, the order of g is unchanged.
>
>So it's enough to prove it for the 2-generator group <g,h>.
>
>quasi

Ok, I can prove it now, but I'll try not to ruin your fun by giving it
all away.

I'll give one general hint, based on my prior reply.

(1) Since the group <g,h> is all that you have to look at, that
suggests some trying to find some simple identity in the symbols g,h.

Here are some further hints, but use them only if needed since they
are fairly strong:

(2) Invert the relationship to see what it says.

(3) Conjugate the original relationship one more time, so now you are
conjugating by h^2 instead of h.

(3) Comparing the results in (1) and (2), h^2 commutes with what?

(4) If h^2 commutes with something, so does any element of <h^2>

(5) But if h has odd order then <h^2>=<h>.

Hmmm. I think I gave away too much. If you looked at these hints,
erase them from memory and retry without them.

quasi
.

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