Let $M:G\to H$ be a homomorphism and let $x$ be in $G$. Suppose that $x$ has order $k$. Show that if $M$ is injective, then $|M(x)|=|x|$

Question

Let $M:G\to H$ be a homomorphism and let $x$ be in $G$. Suppose that $x$ has order $k$. Show that if $M$ is injective the order of $M(x)$ equals order of $x$.

My approach:

$M(x)^k=M(x^k)=M(e)=e'$, where e and e' are identity elements in $G$ and $H$ respectively. Then $|M(x)|$ divides $k$. $(|M(x)| \leq k)$

I'm stuck here, I tried to apply pigeonhole principle, but that seems to lead to nowhere.

Answer 1

Now use injectivity: if $M(x)^p=e'$ for $1\leqslant p\lt k$, then $M(x^p)=e'$ hence by injectivity of $M$, $x^p=e$ which is not possible.

Answer 2

If you know that the order of $x$ is the number of elements in $\langle x\rangle=\{x^n:n\in\mathbb{Z}\}$ (which is how I usually define the order), then

$$ \langle M(x)\rangle=\{M(x)^n:n\in\mathbb{Z}\}= \{M(x^n):n\in\mathbb{Z}\}=M(\langle x\rangle) $$ (the image of $\langle x\rangle$ under $M$). Since $M$ is injective, it preserves the cardinality of subsets.