[sputum]

(Original post by Maths boy)
yes thats exactly what i meant

it's just composing and inverting permutations
so first we knock up pq (check in your text whether it works left-to-right or right-to-left, most go right to left)

(132)(4576)*(15734)(26)
start with 1
(1
q sends 1 to 5 and p sends 5 to 7
(17
q sends 7 to 3 and p sends 3 to 2
(172
q sends 2 to 6 and p sends 6 to 4
(1724
q sends 4 to 1 and p sends 1 to 3
(17243
q sends 3 to 4 and p sends 4 to 5
(172435
q sends 5 to 7 and p sends 7 to 6
(1724356
just to check, q sends 6 to 2 and p sends 2 to 1
(1724356)

try doing sr like this

(Original post by Maths boy)
I think i get it now, I get the inverse as H as it does not change anything it multiplies. just to confirm i am finding the inverses right, is the inverse of a, m?

identity as H yes, inverse of a is indeed m

(Original post by Maths boy)
Done part a, and pretty sure ive done part b, what about part c and d?

The order of an element is how many times you have to compose it with itself to get back to the identity element.
so a.a = c, c.a = something, something.a = ... until you get an = e (aka = H)

once you've done part c part d should follow

[aznkid66]

(Original post by Maths boy)
I think i get it now, I get the inverse as H as it does not change anything it multiplies. just to confirm i am finding the inverses right, is the inverse of a, m?

By the way, I realized that my method was pretty inefficient. You should look for H's row-by-row. For example, the inverse of A is the column that has the element H in row A, aka column M. This allows you to find the inverses of A, B, C, D,... in that order rather than randomly.

---

(Original post by sputum)
The order of an element is how many times you have to compose it with itself to get back to the identity element.
so a.a = c, c.a = something, something.a = ... until you get an = e (aka = H)

Just for clarification, is the order of the identity (in this case H) 0 or 1?
Also, is there any shortcut, like using the order of one element to find the order of another? Or is the fastest way to do each element independently?

[sputum]

(Original post by aznkid66)
Just for clarification, is the order of the identity (in this case H) 0 or 1?
Also, is there any shortcut, like using the order of one element to find the order of another? Or is the fastest way to do each element independently?

I like to think of it as 1 but I'm not sure what the official designation is to be honest. You don't really need to use the value of the order of the identity for anything I've come across

The fastest way in this particular question is to use the group description

"Below is the group table for the symmetries of a regular hexagon, (clockwise rotations of 60o, 120o, 180o, 240o and 300o, reflections in the mid-points of opposite sides and reflections in the lines through opposite vertices but not in this order)"

so the smallest non-trivial rotation will have order 6, the 120o order 3, the 180o order 2 etc
the reflections will intuitively all have order 2
so scamper down the diagonal then pick up the stragglers

If you wanted to do it group-theoretically you could divide the elements by conjugacy classes as each element of a conjugacy class has the same order (but elements of the same order can be in different conjugacy classes so the reverse doesn't work)
and from this it should make sense that the identity is in a conjugacy class by itself (whether it has order labelled 0 or 1)

I can go through conjugating the group if it would help.

[Maths boy]

[QUOTE=sputum;38459024]it's just composing and inverting permutations
so first we knock up pq (check in your text whether it works left-to-right or right-to-left, most go right to left)

(132)(4576)*(15734)(26)
start with 1
(1
q sends 1 to 5 and p sends 5 to 7
(17
q sends 7 to 3 and p sends 3 to 2
(172
q sends 2 to 6 and p sends 6 to 4
(1724
q sends 4 to 1 and p sends 1 to 3
(17243
q sends 3 to 4 and p sends 4 to 5
(172435
q sends 5 to 7 and p sends 7 to 6
(1724356
just to check, q sends 6 to 2 and p sends 2 to 1
(1724356)

try doing sr like this



identity as H yes, inverse of a is indeed m




The order of an element is how many times you have to compose it with itself to get back to the identity element.
so a.a = c, c.a = something, something.a = ... until you get an = e (aka = H)



I can see how q send 1 to 5 and q sends 5 to 7, but how does that make 17?

[aznkid66]

(Original post by Maths boy)

(Original post by sputum)
(132)(4576)*(15734)(26)
start with 1
(1
q sends 1 to 5 and p sends 5 to 7
(17

I can see how q send 1 to 5 and q sends 5 to 7, but how does that make 17?

Note that it's not the number 17 but the elements 1 and 7 in that order.
So really the question you're asking is "Why do we start with 1?" and "Why does 7 result from pq on the previous element (1)?" Are you sure you don't know the answers to these questions?

[sputum]

(Original post by Maths boy)
I can see how q send 1 to 5 and p sends 5 to 7, but how does that make 17?

because p.q(1) = p(5) = 7
so the composite permutation begins (17...)
then see what p.q(7) is and so on

[Maths boy]

(Original post by aznkid66)
Note that it's not the number 17 but the elements 1 and 7 in that order.
So really the question you're asking is "Why do we start with 1?" and "Why does 7 result from pq on the previous element (1)?" Are you sure you don't know the answers to these questions?

Yer im just a bit slow to these things

[Maths boy]

(Original post by sputum)
it's just composing and inverting permutations
so first we knock up pq (check in your text whether it works left-to-right or right-to-left, most go right to left)

(132)(4576)*(15734)(26)
start with 1
(1
q sends 1 to 5 and p sends 5 to 7
(17
q sends 7 to 3 and p sends 3 to 2
(172
q sends 2 to 6 and p sends 6 to 4
(1724
q sends 4 to 1 and p sends 1 to 3
(17243
q sends 3 to 4 and p sends 4 to 5
(172435
q sends 5 to 7 and p sends 7 to 6
(1724356
just to check, q sends 6 to 2 and p sends 2 to 1
(1724356)

try doing sr like this



identity as H yes, inverse of a is indeed m




The order of an element is how many times you have to compose it with itself to get back to the identity element.
so a.a = c, c.a = something, something.a = ... until you get an = e (aka = H)

once you've done part c part d should follow

Ive done it for sr, i get
1453

Start with 1
Sr
1
R takes 1 to 7 and s takes 7 to 4
14
R takes 4 to 1 and s takes 1 to 5
145
R takes 5 to 4 and s takes 4 to 3
1453
R takes 3 to 3 and s takes 3 to 1
(1453)

Is that right? if so what is the next step

[Maths boy]

(Original post by Maths boy)
Thanks man, I have finally figured out how to do it:-), still some things i need help with.

Below is the group table for the symmetries of a regular hexagon, (clockwise rotations of 60o, 120o, 180o, 240o and 300o, reflections in the mid-points of opposite sides and reflections in the lines through opposite vertices but not in this order)
A B C D F G H J K L M N

A C G J M B K A D N F H L

B F H L K A M B N D C G J

C J K D H G N C M L B A F

D M L H C N F D A B K J G

F L M N G H D F K J A B C

G B A F N C H G L M J K D

H A B C D F G H J K L M N

J D N M A K L J H F G C B

K G C B L J A K F H D N M

L N D K B M J L G C H F A

M H F A J L B M C G N D K

N K J G F D C N B A M L H

a) Find the identity
b) Find the inverse of each element
c) Find the order of each of A, J and N
d) State with reasons which of the elements are reflections and which are rotations. For the ones which are rotations describe which rotation is represented by which element. For the ones which are reflections you are not asked to give any details. Note there is more than one correct answer for this question.
e) State the possible orders of the proper subgroups.
f) Find a subgroup of each of the possible (proper) orders and draw up their group tables.
g) For each subgroup decide if it is cyclic and /or Abelian
h) For the subgroup of order 3, find its left cosets and its right cosets. Is the subgroup normal?

I have done part a to c, i am stuck on part d, any help?

[BabyMaths]

(Original post by Maths boy)
I have done part a to c, i am stuck on part d, any help?

What happens when you do the same reflection twice?

[sputum]

(Original post by Maths boy)
Ive done it for sr, i get
1453

Start with 1
Sr
1
R takes 1 to 7 and s takes 7 to 4
14
R takes 4 to 1 and s takes 1 to 5
145
R takes 5 to 4 and s takes 4 to 3
1453
R takes 3 to 3 and s takes 3 to 1
(1453)

Is that right? if so what is the next step

It looks good. Next step is to take an element not already used
so
(1453)(2
any element that goes to itself is left out, although you can leave it in as (x) if it helps.

[Maths boy]

(Original post by BabyMaths)
What happens when you do the same reflection twice?

U get the original thing, cause, if u reflect x, u get reflection of c, if reflect it again u x.

[BabyMaths]

(Original post by Maths boy)
U get the original thing, cause, if u reflect x, u get reflection of c, if reflect it again u x.

So the reflections should be easy to find in the table?

[Maths boy]

yes