Order of Groups: Proving ord(\theta(x)) = ord(x)

[smoothman]

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{r \geq 1: x^r = 1}

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

My answer: Please verify
If \theta(x) = {x^r: r \epsilon Z} then ord(\theta(x)) = 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]" where [G] is an index of H in G, an integer.
So order for any xEG divides order of the group. So ord(\theta(x)) = ord(x)


any suggestions or changes please? thnx :)

 

[HallsofIvy]

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{r \geq 1: x^r = 1}

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

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

This makes no sense. \theta(x) 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]" where [G] 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(\theta(x)) = ord(x)


any suggestions or changes please? thnx :)

Seems to me you could just use the fact that, for any injective homomorphism, \theta, \theta(x^r)= [\theta(x)]^r and \theta(1_G)= 1_H.

 

[smoothman]

i believe we have to show 2 things:

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

ok so basically:

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

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

 

[masnevets]

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