Homomorphisms, Order, Injectivity

Let $G$ and $H$ be groups and $\varphi : G \rightarrow H$ be a homomorphism.

Prove that if $g \in G$ has finite order $n$, then $\varphi(g)$ has finite order in H whose order divides $n$.

By definition of order, $g^n = e$. We can apply the homomorphism to get $\varphi(g^n) = e$ $\implies$ $\varphi(g)^n = e$. Since we have found a positive integer $n$ such that $\varphi(g)^n = e$, we know $\varphi(g)$ has a finite order. Let $m$ be the order of $\varphi(g)$, the smallest positive integer where $\varphi(g)^m = e$.

We can use the division algorithm to rewrite the order $n$ with the order $m$ as $n = mq + r$ where $q, r \in Z$ for $0 \leq r < m$. So then $\varphi(g)^{mq+r} = e$ $\implies$ $[\varphi(g)^m]^q \varphi(g)^r$ = $e^q\varphi(g)^r$ = $\varphi(g)^r$ = e. The division algorithm says that for any integer $n$ and positive integer $m$, there exist unique integers $q$ and $r$ such that $n = mq + r$ for $0 \leq r < m$.

However we said $m$ was the order of $\varphi(g)$ which is the smallest positive integer such that $\varphi(g)^m$. Therefore, r must be 0. So then $n = mq$, and by definition of divisibility, $m | n$. Therefore we have proven that $\varphi(g)$ has finite order in $H$ whose order divides $n$.

Prove that if $\varphi$ is injective, then $|g| = |\varphi(g)|$ for every $g \in G$.

Since this questions asks for all $g \in G$, we must prove two cases: when the order is finite and when the order is infinite. We start with the first case when the order of $g$ is finite.

First, we show that if $\varphi$ is injective, then $ker(\varphi) = \{e_G\}$. Suppose $k \in ker(\varphi)$. By definition of the kernel, $\varphi(k) = e_H$ and $\varphi(e_G) = e_H$. Since $\varphi$ is injective, if $\phi(k) = \phi(e_G)$, then $k = e_G$. So then we have shown the only element in the kernel is the identity element $e_G$.

Suppose $\varphi(g)^m = e_H$ $\implies$ $\varphi(g^m) = e_H$. Since the only element in the kernel is the identity element $e_G$, we know that $g^m$ = $e_H$. Since $|g|$ = $n$ and $g^m = e$, by a similar division algorithm argument as the previous proof, we know $n | m$. In our previous proof, we showed $m|n$. So then $m=n$, and then |$\varphi(g)$| = $|g|$ when the order of $g \in G$ is finite.

Now, we focus on the second case when the order of $g$ is infinite. Then, we want to show the order of $\varphi(g)$ is infinite. Assume for contradiction, that the order of $\varphi(g)$ is finite. Then $\varphi(g)^m = e_H$ $\implies$ $\varphi(g^m) = e_H$. Since $\varphi$ is injective, and we know the only element in the kernel is $e_G$, we know $g^m = e_H$. However, we said that the order of $g$ is infinite. This is a contradiction, so then the order of $\varphi(g)$ must be infinite when the order of $g$ is infinite.

Since we have shown that $|g| = |\varphi(g)|$ when the order of g is infinite or finite, we have shown this statement holds for every $g \in G$.