Order of $\langle f(g)\rangle$ If $f$ is a homomorphism from $G$ to $H,$ if $g\in G$,
in general do $\langle g\rangle$ and $\langle f(g)\rangle$ have the same order? I feel like they can have different orders, but I can't think of a counterexample...
 A: A counterexample: Let $G$ be any non-trivial group and $H=\{1\}$. $f:G\longrightarrow H$ given by $f(g)=1$ is trivially a homomorphism. There exists $x\in G$ which order $n$ is greater than 1, then $|\langle x\rangle|=n$. However, $f(x)=1$ which have order 1, thus $|\langle f(x)\rangle|=1 < |\langle x\rangle |=n$.
A: Following William's comment, we can show that the order of $\left\langle f(g)\right\rangle$ divides the order of $\langle g\rangle$, assuming that both of these have finite order. However, I'd follow his advice of trying this on your own first.
Let $F: \langle g\rangle \to \langle f(g)\rangle$
be the restriction of $f$ to $\langle g\rangle$. That is, $F(x) = f(x)$. $F$ is surjective (why?). By the First Isomorphism Theorem, we know that
$$\langle g\rangle/\ker(F) \cong \langle f(g)\rangle.$$
Since $\langle g\rangle$and $\langle f(g)\rangle$ are finite, the above statement implies the following equality
$$\frac{\mid\langle g\rangle  \mid}{\mid\ker(F)\mid} = \mid\langle f(g)\rangle\mid. $$
We can rearrange this to show that $\mid \langle f(g)\rangle\mid$ divides $\mid\langle g\rangle\mid$. This also shows that $\langle g\rangle$ and $\langle f(g)\rangle$ have the same order iff $f$ is injective when restricted to $\langle g\rangle$.
A: It's a good question.
Actually all that are needed is a few very basic ingredients (more basic than Lagrange and the first isomorphism theorem):

*

*$x^n=e\implies \lvert x\rvert \mid n$ (this may be the trickiest,  you need the division algorithm)

*$h$ a homomorphism $\implies h(x^n)=h(x)^n$

*$x^{\lvert x\rvert}=e$

*$h$ a homomorphism $\implies h(e)=e$ (follows from the second bullet btw).

Now try putting these together to prove $$\lvert h(x)\rvert \mid \lvert x\rvert $$.
In general, we don't get equality (whenever the kernel of $h$ is non-trivial).
