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.