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elementary group theory exercise

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Derek Holt

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Jun 12, 2007, 9:24:02 AM6/12/07
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Despite working in group theory for more years than I care to
remember, I have never come across the elementary exercise below until
a day or two ago.

Show that every left coset in a group G is also a right coset.

The proof is very short.

Derek Holt.

Dave L. Renfro

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Jun 12, 2007, 9:51:16 AM6/12/07
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Derek Holt wrote:

Since a subgroup is normal iff the collection of its
left cosets is equal to the collection of its right
cosets (I believe this is actually how it's defined
in Herstein's "Topics in Algebra", and then he goes
on to prove the seemingly stronger "pointwise result"
that H is normal iff for each g in G the left coset gH
is equal to the right coset Hg), I take it you're
saying that a subset H is normal iff every right
coset is a left coset? If so, that would be a strange
asymmetry. Or am I overlooking something?

Dave L. Renfro

Jack Schmidt

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Jun 12, 2007, 10:13:01 AM6/12/07
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A right coset Hg of H is a left coset gK of a different subgroup K.
The formula for K shows the relation to normality.

Dave L. Renfro

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Jun 12, 2007, 10:30:44 AM6/12/07
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Dave L. Renfro wrote (in part):

>> Or am I overlooking something?

Jack Schmidt wrote:

> A right coset Hg of H is a left coset gK of a
> different subgroup K. The formula for K shows
> the relation to normality.

Ooops, so I was overlooking something. There are
two parameters involved in collections of cosets,
the set H and the element g. The original poster
(Derek Holt) didn't assert the more precise statement
"every left coset of H is a right coset of H", but
rather the less precise statement "every left coset
is a right coset".

Dave L. Renfro

Derek Holt

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Jun 12, 2007, 10:55:30 AM6/12/07
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Yes, and by being less precise in that way, the statement becomes
correct!

I came across this problem because I was trying to understand some
subsets of a group that had the form hHk for a subgroup H and elements
h and k of G.

It took me far too long to realize that hHk = (hHh^--1)(hk), and so
subsets of this form are just cosets of a different subgroup.

Derek Holt.

bluelabel

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Jun 12, 2007, 12:21:25 PM6/12/07
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"Derek Holt" <ma...@warwick.ac.uk> wrote:


ok, but hHk is the left coset of what subgroup?


Arturo Magidin

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Jun 12, 2007, 1:27:18 PM6/12/07
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In article <466ec80e$0$17951$4faf...@reader1.news.tin.it>,
bluelabel <jhsbadfb@#jjkhasdf.it> wrote:
>
>"Derek Holt" <ma...@warwick.ac.uk> wrote:

[...]

"Prove that every left coset in a group G is also a right coset" (or
words to that effect).


>> Yes, and by being less precise in that way, the statement becomes
>> correct!
>>
>> I came across this problem because I was trying to understand some
>> subsets of a group that had the form hHk for a subgroup H and elements
>> h and k of G.
>>
>> It took me far too long to realize that hHk = (hHh^--1)(hk), and so
>> subsets of this form are just cosets of a different subgroup.
>
>
>ok, but hHk is the left coset of what subgroup?

The subgroup hHh^{-1}, as noted above: it is the right coset of
hHh^{-1} determined by the element hk.

The original problem was:

Given a group G, a subgroup H, and an element g in G, show that
there exist an element g' in G, and a subgroup K of G (which may
depend on H and g) such that gH = Kg' as sets.

And the solution is indeed simple: note that for each h in H,

gh = (ghg^{-1})g

thus gH = (gHg^{-1})g.

So we can let K = gHg^{-1} and g' = g. The subgroup K will vary
depending on g (or rather, depending on the left congruence class of g
modulo the normalizer of H in G); in particular, if H is normal, K=H
for all g.

--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================

Arturo Magidin
magidin-at-member-ams-org

Ken Pledger

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Jun 12, 2007, 7:51:53 PM6/12/07
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In article <1181654642....@i38g2000prf.googlegroups.com>,
Derek Holt <ma...@warwick.ac.uk> wrote:

> ....

> Show that every left coset in a group G is also a right coset.

> ....


Here's a simple consequence. Let C be the set of all cosets (left
and right) of all subgroups of G, together with the empty set. Then the
intersection of each collection of members of C is also a member, so in
the usual way one can define a "coset closure" for each subset of G. In
general it's smaller than the subgroup generated by the same subset.
I'm not sure whether this concept is useful, but many years ago I used
it in an exam question. :-)

Ken Pledger.

Keith Ramsay

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Jun 13, 2007, 3:47:52 AM6/13/07
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On Jun 12, 5:51 pm, Ken Pledger <ken.pled...@mcs.vuw.ac.nz> wrote:
| Here's a simple consequence. Let C be the set of all cosets
(left
|and right) of all subgroups of G, together with the empty set. Then
the
|intersection of each collection of members of C is also a member, so
in
|the usual way one can define a "coset closure" for each subset of G.
In
|general it's smaller than the subgroup generated by the same
subset.
|I'm not sure whether this concept is useful, but many years ago I
used
|it in an exam question. :-)

If S is a subset of G and has an element g, then
the coset generated by S is g <g^{-1}S>, where
<g^{-1}S> is the subgroup generated by g^{-1}S
={g^{-1}s : s in S}, since any coset containing
1 is a subgroup, and conversely, and because
cosets are taken to cosets by multiplication by
fixed elements of G. It seems sort of funny that
this should be independent of g, but it is.

I thought the original exercise was a good
inside joke. :-) I was also thinking at first,
"Doesn't that mean it's normal?", but then
it occurred to me, "doesn't that mean _what_
is normal?!"

Keith Ramsay

bluelabel

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Jun 13, 2007, 2:27:39 PM6/13/07
to

"Arturo Magidin" <mag...@math.berkeley.edu> wrote:

> In article <466ec80e$0$17951$4faf...@reader1.news.tin.it>,
> bluelabel <jhsbadfb@#jjkhasdf.it> wrote:
>>
>>"Derek Holt" <ma...@warwick.ac.uk> wrote:
>
> [...]
>
> "Prove that every left coset in a group G is also a right coset" (or
> words to that effect).
>
>
>>> Yes, and by being less precise in that way, the statement becomes
>>> correct!
>>>
>>> I came across this problem because I was trying to understand some
>>> subsets of a group that had the form hHk for a subgroup H and elements
>>> h and k of G.
>>>
>>> It took me far too long to realize that hHk = (hHh^--1)(hk), and so
>>> subsets of this form are just cosets of a different subgroup.
>>


>>ok, but hHk is the left coset of what subgroup?
>
> The subgroup hHh^{-1}, as noted above: it is the right coset of
> hHh^{-1} determined by the element hk.


Sorry, you know that I am a bit dull, even in the easiest things. Do you
mean that hHk is a left coset of hHh^{-1} ??
I understood the question, but I did not understand what Derek was trying to
show with the relation hHk = (hHh^{-1})(hk). I can only see that the subset
hHk is a right coset of hHh^{-1}).


> The original problem was:
>
> Given a group G, a subgroup H, and an element g in G, show that
> there exist an element g' in G, and a subgroup K of G (which may
> depend on H and g) such that gH = Kg' as sets.
>
> And the solution is indeed simple: note that for each h in H,
>
> gh = (ghg^{-1})g
>
> thus gH = (gHg^{-1})g.
>
> So we can let K = gHg^{-1} and g' = g. The subgroup K will vary
> depending on g (or rather, depending on the left congruence class of g
> modulo the normalizer of H in G); in particular, if H is normal, K=H
> for all g.

Yes, I realized what the problem was about.
I understand this, but I fail to replicate your reasoning with hHk =
(hHh^{-1})(hk).
Simply I do not understand if Derek was showing his solution to the problem
or he was making a remark, or something else.


Arturo Magidin

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Jun 13, 2007, 2:58:59 PM6/13/07
to
In article <4670372a$0$4791$4faf...@reader4.news.tin.it>,

bluelabel <jhsbadfb@#jjkhasdf.it> wrote:
>
>"Arturo Magidin" <mag...@math.berkeley.edu> wrote:
>
>> In article <466ec80e$0$17951$4faf...@reader1.news.tin.it>,
>> bluelabel <jhsbadfb@#jjkhasdf.it> wrote:
>>>
>>>"Derek Holt" <ma...@warwick.ac.uk> wrote:
>>
>> [...]
>>
>> "Prove that every left coset in a group G is also a right coset" (or
>> words to that effect).
>>
>>
>>>> Yes, and by being less precise in that way, the statement becomes
>>>> correct!
>>>>
>>>> I came across this problem because I was trying to understand some
>>>> subsets of a group that had the form hHk for a subgroup H and elements
>>>> h and k of G.
>>>>
>>>> It took me far too long to realize that hHk = (hHh^--1)(hk), and so
>>>> subsets of this form are just cosets of a different subgroup.
>>>
>
>
>
>
>>>ok, but hHk is the left coset of what subgroup?
>>
>> The subgroup hHh^{-1}, as noted above: it is the right coset of
>> hHh^{-1} determined by the element hk.
>
>
>Sorry, you know that I am a bit dull,

Not at all. I am dull for not seeing the "left" on your question!

> even in the easiest things. Do you
>mean that hHk is a left coset of hHh^{-1} ??
>I understood the question, but I did not understand what Derek was trying to
>show with the relation hHk = (hHh^{-1})(hk). I can only see that the subset
>hHk is a right coset of hHh^{-1}).

Yes. And now we can do what I did below. If H is a subgroup of G, and
g in G, then the right coset Hg is equal to the left coset gK, where K
= g^{-1}Hg.

Thus, to go from the right coset (hHh^{-1})(hk) to a right coset, we
just need to consider

(hk)(hk)^{-1}(hHh^{-1})(hk) = (hk) (k^{-1}h^{-1}Hh^{-1}hk)
= (hk) k^{-1}Hk.

Thus, the set hHk is a left coset of hHh^{-1}, and a right coset of
k^{-1}Hk, in both cases that coset determined by the element hk.

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