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Orders of Groups

by smoothman
Tags: groups, orders
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smoothman
#1
Jan12-08, 06:10 PM
P: 39
Hi i have completed the answer to this question. Just need your verification on whether it's completely correct or not:

Question:
If G is a group and xEG we define the order ord(x) by:
ord(x) = min{[itex]r \geq 1: x^r = 1[/itex]}

If [itex]\theta[/itex]: G --> H is an injective group homomorphism show that, for each xEG, ord([itex]\theta(x)[/itex]) = ord(x)

My answer: Please verify
If [itex]\theta(x)[/itex] = {[itex]x^r: r \epsilon Z[/itex]} then ord([itex]\theta(x)[/itex]) = ord(x).

For any integer r, we have x^r = e (or 1) if and only if ord(x) divides r.

In general the order of any subgroup of G divides the order of G. If H is a subgroup of G then "ord (G) / ord(H) = [G:H]" where [G:H] is an index of H in G, an integer.
So order for any xEG divides order of the group. So ord([itex]\theta(x)[/itex]) = ord(x)


any suggestions or changes please? thnx :)
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HallsofIvy
#2
Jan12-08, 08:32 PM
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P: 39,348
Quote Quote by smoothman View Post
Hi i have completed the answer to this question. Just need your verification on whether it's completely correct or not:

Question:
If G is a group and xEG we define the order ord(x) by:
ord(x) = min{[itex]r \geq 1: x^r = 1[/itex]}

If [itex]\theta[/itex]: G --> H is an injective group homomorphism show that, for each xEG, ord([itex]\theta(x)[/itex]) = ord(x)

My answer: Please verify
If [itex]\theta(x)[/itex] = {[itex]x^r: r \epsilon Z[/itex]} then ord([itex]\theta(x)[/itex]) = ord(x).
This makes no sense. [itex]\theta(x)[/itex] is a single member of H, not a set of members of G.

For any integer r, we have x^r = e (or 1) if and only if ord(x) divides r.

In general the order of any subgroup of G divides the order of G. If H is a subgroup of G then "ord (G) / ord(H) = [G:H]" where [G:H] is an index of H in G, an integer.
An "index of H in G"? It is not said here that H has to be a subset of G!

So order for any xEG divides order of the group. So ord([itex]\theta(x)[/itex]) = ord(x)


any suggestions or changes please? thnx :)
Seems to me you could just use the fact that, for any injective homomorphism, [itex]\theta[/itex], [itex]\theta(x^r)= [\theta(x)]^r[/itex] and [itex]\theta(1_G)= 1_H[/itex].
smoothman
#3
Jan13-08, 07:29 AM
P: 39
i believe we have to show 2 things:

i) [itex](\theta(x))^a = e'[/itex]
ii) [itex]0 < b < a \implies (\theta(x))^b \neq e'[/itex].

ok so basically:

If ord(x)=a then [itex]\left[ {\phi (x)} \right]^a = \left[ {\phi (x^a )} \right] = \phi (e) = e'[/itex].
Now suppose that [itex]ord\left[ {\phi (x)} \right] = b < a[/itex].
Then
[itex]\left[ {\phi (x)} \right]^b = e' = \left[ {\phi (x)} \right]^a [/itex]
[itex]\phi (x^b ) = \phi (x^a )[/itex]
[itex]x^b = x^a[/itex] (injective)
[itex]x^{a - b} = e[/itex]

there seems to be a contradiction where if [itex]x^{a} = x^{b}[/itex], then [itex]\left[ {\phi (x)} \right]^b = e'[/itex] which is not what statement (ii) says.
am i correct in this assumption? any ideas on how to deal with this?

masnevets
#4
Jan16-08, 09:12 PM
P: 49
Orders of Groups

When you have [itex]x^{a-b} = e[/itex], then a-b is positive since you assumed b<a. But this contradicts the definition of order. Done.


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