Re: homomorphism from D6 to S5 or A5

In article <29623761.1184784931146.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
Mathematicien <achnine0@xxxxxxxxxxx> wrote:
In article
<16938381.1184436790774.JavaMail.jakarta@xxxxxxxxxxxxx
forum.org>,

is there an injective homomorphism from D6 to S5 !?

[...]

Okay, you quoted. Good. Now, proceed to the next step: trim your
responses a bit by removing stuff that is not important. In
particular, you should almost always remove the .sig from the post you
are replying to.

about
D6 = < r,s | r^12 = s^2 = 1, sr = r^5s >.

This is incorrect (my mistake): it should be

D6 = < r,s | r^6 = s^2 = 1, sr = r^5s >.

I have seen another 'definition' of D6 on wolfram mathworld. Are there more definitions? Is the definition of in general:
Dn = < r,s | r^(2n) = s^2 = 1, sr = r^(n-1)s > ?

In the convention you are following, it should be:

Dn = < r,s | r^n = s^2 = 1, sr = r^(n-1)s >.

s and r are generators of Dn. How did one choose these definitions?

Take the regular n-gon inscribed in the unit disk, with one vertex at
(1,0).

Dn is the group of all rigid symmetries of this n-gon.

The element r corresponds to a rotation in the positive direction by
360/n degrees; that is, the rotation that moves the vertex that was in
(1,0) to the "next" vertex above it. Repeating it n times will lead to
the identity. This tells you that r^n = 1.

The element s corresponds to the reflection about the
x-axis. Repeating it twice yields the identity. This tells you that
s^2 = 1.

That these two operations yield all possible rigit motions of the
n-gon is not too hard: any rigid motion is completely determined by
what it does to two adjacent vertices; so it is enough to know where
the vertex from (1,0) ends up, and whether the vertices end up in the
same or opposite ordering you started with. This amounts to not doing
the reflection, or doing it; and following it up by a suitable
rotation. So every rigid transformation can be written as either r^k
or r^ks for some integer k, 0<= k < n.

Finally, if you first do the rotation and then the reflection, you end
up with the same thing as if you first do the reflection and then
repeat the rotation n-1 times. This tells you that sr = r^(n-1)s
(think of the elements as functions, so you compose them right to
left; "sr" means "first do r, then do s").

That these three rules are enough to describe the group is not too
hard now: any element of the group can be written as a product of r's
and s's. The rule sr = r^(n-1)s allows you to "move" all the r's over
to the left, so you end up with something of the form r^k s^m. Now use
the fact that r^n = 1 to reduce k to a number 0<= k < n, and use the
fact that s^2 = 1 to reduce m to either 0 or 1, depending on whether m
is even or odd.

about
how many normal subgroups of A5 are there?
I think 2: A5 and {1}. I dont have to use this yet, because it will be taught (in the next chapter of our syllabus).
Any way: would you like to give me some examples of congugacy classes
of some element in Sn which split into 2 classes in An?

You mean, would I like to do your homework for you? No, not
particularly.

How about you try a few with a small n? You can do n=4 and n=5 by
hand, surely.

and other
examples where this does not happen?

->I<- already passed my group theory courses. I don't have to do your
homework for you.

--
======================================================================
"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

.

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