Prove $f$ injective implies $o(f(g)) = o(g)$.

Question

Suppose $G_1,G_2$ are groups and $f:G_1 \to G_2$ is a homomorphism. Prove $f$ injective implies $o(f(g)) = o(g) \space\forall g\in G$.

Suppose $g\in G , g'\in G$ such that $o(g)=n, f(g)=g'$. Then

$$\begin{align} e&=f(e)\\ &=f(g^n)\\ &=(f(g))^n\\ &=(g')^n, \end{align}$$

which implies $o(g')\mid n.$

I can't see how using the injectivity helps me find that $o(f(g)) = o(g)$.

Help please.

Answer 1

The group

$$\langle g\rangle=\{e,g,\dots, g^{n-1}\}$$

has $n$ distinct elements, since $|g|=n$.

Consider $g'^i=g'^j$ for some $i,j\in\{0,1,\dots, n-1\}$. Here $g'=f(g)$, so

$$\begin{align} g'^i=g'^j&\implies f(g)^i=f(g)^j\\ &\implies f(g^i)=f(g^j)\quad(f\text{ is a homomorphism})\\ &\implies g^i=g^j\quad(f\text{ is injective})\\ &\implies i=j\quad(\text{these powers of }g\text{ are distinct}). \end{align}$$

Therefore, the elements of $\{e,g',\dots, g'^{n-1}\}$ are distinct.

But $|f(g)|\mid |g|$.

Clearly, if $|g|$ is infinite, then so is $f(g)$ by injectivity, for suppose otherwise: then $|f(g)|=m$ would imply $f(g)^m=f(g)^{2m}$, which would mean $f(g^m)=f(g^{2m})$, giving $g^m=g^{2m}$ by injectivity; but then $|g|$ would be finite.

Thus $|f(g)|=|g|$.

Answer 2

Since the $n$ elements of $\langle g\rangle=\{e_1,g,g^2,\dots,g^{n-1}\}$ are pairwise distinct, then by the injectivity of $f$ so are those of the set $\{f(e_1)=e_2,f(g),f(g^2),\dots,f(g^{n-1})\}=\{e_2,f(g),f(g)^2,\dots,f(g)^{n-1}\}$. You have already proved that $o(f(g))\mid n$, and now we have proved that there are $n$ distinct powers of $f(g)$. Therefore, $o(f(g))=n$.

Answer 3

Now let $o(g')=k$. Then $$\begin{align*} e &= (g')^k\\ &= (f(g))^k\\ &= f(g^k). \end{align*}$$ By injectivity of $f$, $g^k=e$, so $n\mid k$.

You can make the argument work for infinite order without having to divide into parts by noting that your original argument and this one show that for any integer $k$, $$ g^k =e\iff (f(g))^k=e.$$ Since the order of $x$ is defined as the nonnegative generator of the ideal of integers $k$ for which $x^k=0$, this proves $g$ and $f(g)$ have the same order.

Answer 4

Here is an alternative proof to the ones presented so far (although in essence all such proofs are the same).

Let $f: G_1 \to G_2$ be injective, and $g \in G_1$. Let $H = \langle g \rangle \leq G_1$. Then $f$ restricts to an injective map $h = f|_H : H \to G_2$, so since $f|_H(H) = \langle h(g) \rangle$,

$$|\langle h(g) \rangle | = |f|_H(H)| = |H| = |\langle g \rangle|$$

Therefore $\text{ord}(f(g)) = \text{ord}(g)$.