Question

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Answer 1

I know that there is at least one element of order 2, and I know that every element has order 1,2,3,6. I cant use Lagrange theorem.

Answer 2

So I need to show that it is not the case that every non trivial element has order 2.

Answer 3

i guess you are not allowed to use any theorem right? just have to do it by brute force?

Answer 4

for one thing, there are exactly three groups of order 6
each has an element of order 3

Answer 5

yeah, I know that we cant have an element of order > 3
I know that if there is any amount of order two elements there must be an odd amount
we know that there is at least one element of order 2

Answer 6

there are only 2 groups of order 6

Answer 7

Z_6 and S_3

Answer 8

so if there is an element a of order 6 then we have a^2 has order 3. If G has no element of order 6, we suppose by contradiction we have all elements of order 2.

from here I can show it has a subgroup of order 4, and that breaks Lagrange but we cant use it.

I also know that if every non trivial element has order 2 then the group is Abelian.

Answer 9

oh right,
\[\mathbb{Z}_6\simeq \mathbb{Z}_2\times \mathbb{Z}_3\]

Answer 10

hmm, somehow I think I need to break the fact that we would have a commutative group

Answer 11

ok well one thing is for sure
if it is cyclic then it has an element of order 3

Answer 12

because S_3 does not commute

Answer 13

how do I show that?

Answer 14

so i guess you have to start by supposing the group is not cyclic

Answer 15

oh if it is cyclic then it i is generated by something say \(a\) and the elements are \[\{e, a, a^2, a^3, a^4, a^5\}\] and \(a^2\) has order 3

Answer 16

damnt i meant abelian sorry

Answer 17

right, so no element of order 6

Answer 18

so you can assume that it is not cyclic, and then lets see...

Answer 19

so we have left as options 1,2,3 and e is the only element of order 1. So assume that every element has order 2

Answer 20

ok then if every element has order 2, it has to be abelian for sure
that means there is a subgroup of order 4, which is impossible
this is kind of a long way around the obvious, but that is ok

Answer 21

yes but I cant use Lagrange, and that contradiction needs Lagrange.

Answer 22

yikes

Answer 23

but there might be a less obvious contradiction there.

Answer 24

i hate these problems where you cannot use something that makes it obvious

Answer 25

yeah, and its a 500 level abstract class. We just got done constructing extension fields, then we are asked to forget it all and basically start with only the Sudoku property of Cayley tables and the Definition of a group.

Answer 26

good stuff

Answer 27

i will think, but i gotta run
sounds fun
good luck

Answer 28

ty :)

Answer 29

what is a sudoku table? maybe you have to try and write one out since you can't use any theorems?

Answer 30

the problem as i see it is that if all the elements are of order 2, you are going to get stuck after the table for \(\{2, a, b, ab\}\) there will be no more room

Answer 31

Sudoku property just says that row and column of a Cayley table has exactly every element.

Answer 32

that was supposed to be a \(e\) not a \(2\)

Answer 33

oh maybe that is the proof!

Answer 34

right, so if we have a is of order 2, then we have {a,e} then if we have b is of order 2, then we have {e,a,b,ab} (we know its abelian), if we add c which has order 2, then we have {e,a,b,c,ab,ac,cb} and these are all unique

Answer 35

i wrote on my paper \(\{e, a, b, ab, c, d\}\) but i guess it is the same
i just asserted that \(ab=c\) because it has to be something

Answer 36

ok we don't have uniqueness, but we always have odd amount of elements.

Answer 37

G is closed and since \(a^2=b^2=e\) you know \(ab\neq a\) and \(ab\neq b\) so it has to be something, i just called it \(c\)

Answer 38

do we have uniqueness?

Answer 39

Can't we use Cauchy Theorem?
http://en.wikipedia.org/wiki/Cauchy%27s_theorem_%28group_theory%29

Answer 40

lol no we can't use anything, elsewise we would be quite done

Answer 41

nope:(

Answer 42

but i think the table will give it
you will have the complete "sudoku" table filled in with only the first 4 elements, then you cannot complete the table the way you want

Answer 43

ok with the second element we get the set {e,a,b,c,ab, ab, bc, abc} this is an 8 element group.

Answer 44

@zzr0ck3r high fives @satellite73 and walks away.