# Orders of Groups

by smoothman
Tags: groups, orders
 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{$r \geq 1: x^r = 1$} 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: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($\theta(x)$) = ord(x) any suggestions or changes please? thnx :)
Math
Emeritus
Thanks
PF Gold
P: 39,490
 Quote by 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).
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: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($\theta(x)$) = 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 $\theta(1_G)= 1_H$.
 P: 39 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?
 P: 49 Orders of Groups When you have $x^{a-b} = e$, then a-b is positive since you assumed b

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