Artificial intelligent assistant

Proving a function is a monomorphism Suppose that H is a group and there is a group homomorphism $f : D_{14} \to H$ with the property that $f(o) \not= eH$ and $f(r) \not= eH$. Show that $f$ is a monomorphism. Where $o$ is rotation and $r$ is reflection. So what I think I need to show is that $\mathrm{ker}(f) = eD_{14}$ but I'm not sure if there's another way. My process was to try and eliminate all other elements of $D_{14}$ (dihedral group on heptagon) leaving only the identity element. I was using $f(o) \not= eH$ so $f(o)^n$ $ \not= eH^n=eH$ to eliminate them but then I realised that $f(o)^7$ is $f(e)$ and by my logic that shouldn't equal $eH$ so I'm going wrong somewhere. Where am I going wrong? Thanks for any help

The kernel of $f$ is a _normal_ subgroup, and the only normal subgroup of $D_{14}$ that does not contain a rotation is the trivial one.

xcX3v84RxoQ-4GxG32940ukFUIEgYdPy b536133c37ecbec7c2312a9ef1dea942