# order of the group

• Jan 14th 2009, 06:44 AM
knguyen2005
order of the group
Let G be a group and x belongs to G,

Define ord(x) = min{r >= 1 : x^r = 1}

If f: G map to H is an injective group homomorphism. Show that, for
each x in G, ord(f(x)) = ord(x).

I think I come up with the approach is that , I need to show that
ord(f(x)) divides r and hence the result is proved.

Or: I can somehow show that ord(f(x)) = min{ r >= 1 : (f(x))^r = 1} which is by the definition.

Can you tell me whether the above approaches are right or wrong, if they are wrong then show me how to solve this question please?
• Jan 14th 2009, 09:52 AM
ThePerfectHacker
Quote:

Originally Posted by knguyen2005
Let G be a group and x belongs to G,

Define ord(x) = min{r >= 1 : x^r = 1}

If f: G map to H is an injective group homomorphism. Show that, for
each x in G, ord(f(x)) = ord(x).

I think I come up with the approach is that , I need to show that
ord(f(x)) divides r and hence the result is proved.

Or: I can somehow show that ord(f(x)) = min{ r >= 1 : (f(x))^r = 1} which is by the definition.

Can you tell me whether the above approaches are right or wrong, if they are wrong then show me how to solve this question please?

Let $\displaystyle \phi: G\to H$ be an injective homomorphism.
And let $\displaystyle x\in G$ (assuming it has finite order) have order $\displaystyle k$.
Thus, $\displaystyle x^k = e$ but then $\displaystyle \phi(x)^k = \phi(x^k) = e'$.
Say that there is $\displaystyle j<k$ with $\displaystyle \phi^(x)^j = e'$ but then $\displaystyle \phi (x^j) = e' \implies x^j = e$.
But this is contradition because $\displaystyle j<k$ and so $\displaystyle k$ is order of $\displaystyle \phi(x)$.
• Jan 16th 2009, 01:55 PM
knguyen2005
[Thus, $\displaystyle x^k = e$ but then $\displaystyle \phi(x)^k = \phi(x^k) = e'$.
Say that there is $\displaystyle j<k$ with $\displaystyle \phi^(x)^j = e'$ but then $\displaystyle \phi (x^j) = e' \implies x^j = e$.

I am not sure I understand, how come you got: $\displaystyle \phi(x)^k = \phi(x^k) = e'$. And finally when you concluded that $\displaystyle \phi (x^j) = e' \implies x^j = e$. How can you imply this result?

Thank you again for your precious time
• Jan 16th 2009, 09:30 PM
Isomorphism
Quote:

I am not sure I understand, how come you got: $\displaystyle \phi(x)^k = \phi(x^k) = e'$
$\displaystyle \phi(x^k) = \phi(x) ^k$ because $\displaystyle \phi$ is a homomorphism. Since $\displaystyle x^k = e$ and $\displaystyle \phi$ is a group homomorphism, which means it maps identity to identity. Thus $\displaystyle \phi(x)^k = \phi(x^k) = e'$

Quote:

. And finally when you concluded that $\displaystyle \phi (x^j) = e' \implies x^j = e$. How can you imply this result?
He assumed that \phi is injective (or one-one). And $\displaystyle \phi (x^j) = e' = \phi(e)$, thus injectivity forces $\displaystyle \phi (x^j) = \phi(e) \implies x^j = e$.
• Jan 16th 2009, 11:01 PM
ursa
Quote:

I am not sure I understand, how come you got: http://www.mathhelpforum.com/math-he...38419add-1.gif
X Є G
O(G)=n
X^n =e
f(X^n)=f(e)
f(X.X.X.X.X….X)=e’
f(X). f(X) . f(X) . f(X) . f(X)…… . f(X)=e’
(f(X))^n=e’
(f(X))^n=f(X^n)=e’
O(f(X))=n
O(H)=n
O(H)=O(G)
• Jan 17th 2009, 06:47 AM
knguyen2005
Thanks very much you guys. Now I understand it now. It makes sense to me due to your clear explanation. Cheers