1. ## groups

can someone help me with this questuon please its fundamnetal to the topic and preventing me understand the topic!!

Thanks

Edgar

if G is a group and x is an element of G we define the order ord(x) of x by

ord(x) = min {r > or equal to 1 : x^r=1}

if f:G maps to His an injective group homomorphism show that for each x
ord (f(x)) = ord (x)

2. Originally Posted by edgar davids

if G is a group and x is an element of G we define the order ord(x) of x by

ord(x) = min {r > or equal to 1 : x^r=1}

if f:G maps to His an injective group homomorphism show that for each x
ord (f(x)) = ord (x)
These are finite groups, right?
Because if not, how do you know that $\displaystyle x$ has order?

3. yes there are finite groups

4. Originally Posted by edgar davids
yes there are finite groups

Because, $\displaystyle f$ is one-to-one.
Thus, a injective function between two finite sets.
Now, $\displaystyle H\leq G$ thus, $\displaystyle |H|\leq |G|$ (again they are finite).
But you cannot have a injective map from $\displaystyle f:X\to Y$ where they are non-empty and $\displaystyle |X|>|Y|$. Thus, the only possible choice is that $\displaystyle H=G$, that is the subgroup is the group itself. Now a injective map between the same finite non-empty set is also surjective (Pigeonhole principle). Thus, what we really have $\displaystyle f:G\to G$ and it is an automorphism.
Thus, by homomorphism properties,
$\displaystyle [f(x)]^n=f(x^n)=f(e)=e$*
(The reason why $\displaystyle f(e)=e$ is because it is an automorphism).
Thus, we know the order of $\displaystyle f(x)$ is at most $\displaystyle n$. If it was less than $\displaystyle n$ by * we have that $\displaystyle x$ has smaller order which cannot be contrary to our assumption.