Basic group theory and permutations
Stephen Horne
As a programmer, I've been aware for a very long time that abstract
algebra is important, yet somehow managed to avoid learning much about
it. I decided to rectify that, and am currently learning the basics of
group theory. Book/web learning has its problems, and I have a few
uncertainties that are bugging me, so...
If I have a set of n items, I can easily define a set of the n!
permutations of those items. I can also define a group based on this
set, defining the needed operator by defining a bijection to the set
of integers modulo n! and defining my group of permutations to be
isomorphic to the group of integers modulo n! with +.
That is, I can assign each permutation an integer label { 0 .. n!-1 },
and define my associative operator for my set of permutations such
that a.b=c iff f(a)+f(b)=f(c) where f is my bijective function from
the set of permutations to the set of integers modulo n!.
I *hope* that I have my terms right, so this will be understood ;-)
Question is - have I got the following right...
Although this is a group of permutations I have defined, it is *not* a
permutation group. A permutation group, by definition, has an operator
that composes permutations, similar to composition of functions.
In most cases (probably iff n>=3) it is impossible to define a group
isomorphic to [integers modulo n!, +] which is a permutation group.
A permutation group that includes all possible permutations of n items
(in general, probably if n>=3) cannot be cyclic. You cannot choose a
single permutation that generates the whole group through repeated
composition of an element-to-element-bijection-based permutation.
Any group that is isomorphic to [integers modulo n!, +] is cyclic
because all finite cyclic groups are isomorphic to [integers modulo m,
+] where m is the order of the group, so my group of permutations
above must be cyclic where a permutation group (for all permutations
of n>=3 items) cannot be cyclic.
Thus (for n>=3) for my group of permutations to be a permutation group
would be a contradiction.
That said, a permutation group including all permutations of n items
will have cyclic subgroups. Each cyclic subgroup is itself a
permutation group - it just doesn't include all the possible
permutations of the n items. Such a cyclic subgroup cannot contain
more than n permutations in its set (where n is still the number of
items to permute).
So if I were interested in permutations of { a, b, c }, I could define
a group via isomorphism such as...
0 <-> a, b, c
1 <-> b, a, c
2 <-> a, c, b
3 <-> c, a, b
4 <-> b, c, a
5 <-> c, b, a
With the result that, by my groups operator . and its bijection to
modulo 6 +...
[b, a, c] . [a, c, b] = [c, a, b] (because 1 + 2 = 3)
This group is cyclic, and (no matter how I define my isomorphism) it
is not a permutation group.
I can define a permutation group based on a subset of my permutations
such as...
{ [a, b, c], [b, c, a], [c, a, b] }
The bijection for the isomorphism is then...
0 <-> a, b, c (identity)
1 <-> b, c, a (element that generates this cyclic group)
2 <-> c, a, b (alternative element that generates... - inverse)
I can certainly define a permutation group which covers all 6
permutations, and this group will be a cyclic subgroup of that group,
but I cannot define a cyclic permutation group including all 6
permutations. The largest cyclic permutation group I can have based on
these three elements will contain three permutations. I think the
group of three permutations above is the only possibility, though
there is an alternative isomorphism to integers modulo 3, swapping the
roles of [b, c, a] and its inverse [c, a, b]...
0 <-> a, b, c (identity, as above)
1 <-> c, a, b
2 <-> b, c, a
Am I making sense, or have I got the wrong idea?
Christopher Henrich
Stephen Horne <sh006...@blueyonder.co.uk> wrote:
[snipped]
>
> If I have a set of n items, I can easily define a set of the n!
> permutations of those items. I can also define a group based on this
> set, defining the needed operator by defining a bijection to the set
> of integers modulo n! and defining my group of permutations to be
> isomorphic to the group of integers modulo n! with +.
permutations, if n>2. The quickest way to see this is to realize that
except for n=2, the composition of permutations is non-commutative; that
is, the order of the two operands matters to the result. If A and B are
permutations, then AB is usually different from BA.
>
> Question is - have I got the following right...
>
>
> Although this is a group of permutations I have defined, it is *not* a
> permutation group. A permutation group, by definition, has an operator
> that composes permutations, similar to composition of functions.
>
> I can define a permutation group based on a subset of my permutations
> such as...
>
> { [a, b, c], [b, c, a], [c, a, b] }
>
> The bijection for the isomorphism is then...
>
> 0 <-> a, b, c (identity)
> 1 <-> b, c, a (element that generates this cyclic group)
> 2 <-> c, a, b (alternative element that generates... - inverse)
>
> I can certainly define a permutation group which covers all 6
> permutations, and this group will be a cyclic subgroup of that group,
> but I cannot define a cyclic permutation group including all 6
> permutations. The largest cyclic permutation group I can have based on
> these three elements will contain three permutations. I think the
> group of three permutations above is the only possibility, though
> there is an alternative isomorphism to integers modulo 3, swapping the
> roles of [b, c, a] and its inverse [c, a, b]...
>
> 0 <-> a, b, c (identity, as above)
> 1 <-> c, a, b
> 2 <-> b, c, a
>
> Am I making sense, or have I got the wrong idea?
groups can have the same order and not be isomorphic.
--
Christopher J. Henrich
chen...@monmouth.com
http://www.mathinteract.com
"A bad analogy is like a leaky screwdriver." -- Boon
Stephen Horne
<chen...@monmouth.com> wrote:
>I think you have got it right. The upshot of it all is that two finite
>groups can have the same order and not be isomorphic.
Thanks - it's good to have some confidence that I've understood things
so far, before taking the next step.