TITLE: False homework proof?: The image of an element has the same order.
QUESTION [13 upvotes]: I have this assignment from a homework that I'm pretty sure is wrong.  It asks me to prove

Given a group homomorphism $\phi: G\rightarrow G'$, if $g\in G$ has order $k$ then so does $\phi(g)$.

I would think this is intuitively false with an easy counter-example:  The trivial homomorphism $\phi(x)=1$ for all $x\in G$.  Intuitively, you need an isomorphism for the order of an element to be preserved, right?

REPLY [2 votes]: The other answers correctly discuss that the order of $\phi(g)$ must divide the order of $g$, but neglect to mention that some people use the rarer convention that $g$ is of order $k$ simply if $g^k = 1$, even if $k$ isn't the minimal such exponent. Rare as this convention is, the statement you were given to prove is true under it, so it's probably what was meant. Check carefully what convention your book/class uses.